SOLUTION: Please help and show me if I am doing this right! I have to solve an quadratic formula: 3x^=19x-14=0 Ax^+bx+c=0, a=0 x=(-b±√(b^2-4ac))/2a 3x^+19x-14=0 A=3 B=19

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Question 593806: Please help and show me if I am doing this right! I have to solve an quadratic formula: 3x^=19x-14=0
Ax^+bx+c=0, a=0
x=(-b±√(b^2-4ac))/2a
3x^+19x-14=0
A=3
B=19
C=14
3^=(-19±√(〖-19〗^2-4(3)(14)))/(2(3))
3^= (19±√(〖361〗^+ 168))/6
3^= (19±√529)/6
19+23=42=7
6 6
my answer is 7 over 6.
Found 2 solutions by Alan3354, bucky:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Please help and show me if I am doing this right! I have to solve an quadratic formula: 3x^=19x-14=0
Ax^+bx+c=0, a=0
x=(-b±√(b^2-4ac))/2a
3x^+19x-14=0
A=3
B=19
C=14 *********** C = -14
3^=(-19±√(〖-19〗^2-4(3)(14)))/(2(3)) ???? 3^ = ???? It's x =
3^= (19±√(〖361〗^+ 168))/6 ************ You changed signs again, got it right.
3^= (19±√529)/6
19+23=42=7
------------
(-19+23)/6 = 4/6
x = 2/3
--------
(-19-23)/6 = -42/6
x = -7
--------
6 6
my answer is 7 over 6.
------
You were close.
===================
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 3x%5E2%2B19x%2B-14+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%2819%29%5E2-4%2A3%2A-14=529.

Discriminant d=529 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-19%2B-sqrt%28+529+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2819%29%2Bsqrt%28+529+%29%29%2F2%5C3+=+0.666666666666667
x%5B2%5D+=+%28-%2819%29-sqrt%28+529+%29%29%2F2%5C3+=+-7

Quadratic expression 3x%5E2%2B19x%2B-14 can be factored:
3x%5E2%2B19x%2B-14+=+%28x-0.666666666666667%29%2A%28x--7%29
Again, the answer is: 0.666666666666667, -7. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B19%2Ax%2B-14+%29

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Let's begin by ensuring that I interpret your problem correctly. You are asking to solve for the roots of the following equation:
.
3x%5E2+%2B+19x+-14+=+0
.
If that is correct, then here is how you solve it. Begin by comparing it to the standard form of:
.
ax%5E2+%2B+bx+%2B+c+=+0
.
By comparing each term of the standard form to the problem you should see that the following are the constants:
.
a = +3
b = +19 and
c = -14 (note the minus sign)
.
Now that you have the constants you can substitute them into the quadratic solution:
.
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
.
Substituting the above listed values for a, b, and c makes this solution equation become:
.
x+=+%28-%28%2B19%29+%2B-+sqrt%28+%2819%29%5E2-4%2A3%2A%28-14%29+%29%29%2F%282%2A3%29+
.
Notice that the first term in the numerator is -19. Next, within the radical the 19-squared is equal to 361 and the -4 times +3 times -14 is +168. And finally the denominator of +2 times +3 is +6. With these, the equation becomes:
.
x+=+%28-19+%2B-+sqrt%28+361%2B168+%29%29%2F6+
.
Adding the two terms inside the radical sign results in +529 as shown below:
.
x+=+%28-19+%2B-+sqrt%28+529+%29%29%2F6+
.
And since the square root of 529 is 23, the solution equation becomes:
.
x+=+%28-19+%2B-+23%29%2F6+
.
This shows that there are two values for x. One of them occurs when the plus sign precedes the 23 which results in:
.
x+=+%28-19+%2B+23%29%2F6
.
And by adding the two terms in the numerator (-19 + 23) this simplifies to:
.
x+=+4%2F6 which reduces to
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+x+=+2%2F3
.
The second value for x comes when the minus sign precedes the 23 and the equation for x becomes:
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x+=+%28-19+-+23%29%2F6+
.
Combine the -19 and the -23 in the numerator to get -42 so that this time the equation for x is:
.
x+=+%28-42%29%2F6
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And dividing the denominator into the numerator results in the second value for x of:
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x+=+-7
.
So, in summary, the equation has two answers as roots. They are:
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x+=+2%2F3 and x+=+-7
.
Hope this helps you. Your first mistake was that c was equal to -14, not +14 as you said, but somehow you managed to get back on track and came up with the square root of 529. The next error was when you said:
.
3^= (19±√529)/6
.
The left side of this equation should have just been x, and in the numerator of the right side the 19 should have been -19. Other than those errors, you were on the right track.
.