SOLUTION: Solve by using the quadratic formula: 3x^2+4x-15=0

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Question 116773: Solve by using the quadratic formula: 3x^2+4x-15=0
Found 2 solutions by jim_thompson5910, bucky:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let's use the quadratic formula to solve for x:


Starting with the general quadratic

ax%5E2%2Bbx%2Bc=0

the general solution using the quadratic equation is:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29



So lets solve 3%2Ax%5E2%2B4%2Ax-15=0 ( notice a=3, b=4, and c=-15)




x+=+%28-4+%2B-+sqrt%28+%284%29%5E2-4%2A3%2A-15+%29%29%2F%282%2A3%29 Plug in a=3, b=4, and c=-15



x+=+%28-4+%2B-+sqrt%28+16-4%2A3%2A-15+%29%29%2F%282%2A3%29 Square 4 to get 16



x+=+%28-4+%2B-+sqrt%28+16%2B180+%29%29%2F%282%2A3%29 Multiply -4%2A-15%2A3 to get 180



x+=+%28-4+%2B-+sqrt%28+196+%29%29%2F%282%2A3%29 Combine like terms in the radicand (everything under the square root)



x+=+%28-4+%2B-+14%29%2F%282%2A3%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



x+=+%28-4+%2B-+14%29%2F6 Multiply 2 and 3 to get 6

So now the expression breaks down into two parts

x+=+%28-4+%2B+14%29%2F6 or x+=+%28-4+-+14%29%2F6

Lets look at the first part:

x=%28-4+%2B+14%29%2F6

x=10%2F6 Add the terms in the numerator
x=5%2F3 Divide

So one answer is
x=5%2F3



Now lets look at the second part:

x=%28-4+-+14%29%2F6

x=-18%2F6 Subtract the terms in the numerator
x=-3 Divide

So another answer is
x=-3

So our solutions are:
x=5%2F3 or x=-3

Notice when we graph 3%2Ax%5E2%2B4%2Ax-15, we get:

+graph%28+500%2C+500%2C+-13%2C+15%2C+-13%2C+15%2C3%2Ax%5E2%2B4%2Ax%2B-15%29+

and we can see that the roots are x=5%2F3 and x=-3. This verifies our answer

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
.
3x%5E2%2B4x-15=0
.
This is in the standard quadratic form of:
.
ax%5E2+%2B+bx+%2B+c+=+0
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By comparing the standard form to the given problem, you can see that a = 3, b = 4, and c = -15
.
For the standard form, the values of x that satisfy the equation are given by:
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x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
.
So all you have to do to solve the given equation is to substitute 3 for a, 4 for b, and -15
for c into the equation for x. When you do, that equation becomes:
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x+=+%28-%284%29+%2B-+sqrt%28+4%5E2-4%2A3%2A%28-15%29+%29%29%2F%282%2A3%29+
.
Multiply out the denominator 2*3 = 6 to make the equation become:
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x+=+%28-%284%29+%2B-+sqrt%28+4%5E2-4%2A3%2A%28-15%29+%29%29%2F6+
.
Work inside the radical. 4%5E2+=+16 and -4%2A3%2A%28-15%29+=+180 and substituting
these values results in:
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x+=+%28-%284%29+%2B-+sqrt%28+16%2B180%29%29%2F6+
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Combine the terms in the radical:
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x+=+%28-%284%29+%2B-+sqrt%28+196%29%29%2F6+
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But the square root of 196 is 14. Substituting this results in:
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x+=+%28-%284%29+%2B-+14%29%2F6+
.
Note that -(4) = -4 which simplifies the equation to:
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x+=+%28-4+%2B-+14%29%2F6+
.
So there are two possible values of x as follows:
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x+=+%28-4+%2B+14%29%2F6++=+10%2F6+=+5%2F3
.
and
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x+=+%28-4+-+14%29%2F6+=+-18%2F6+=+-3
.
Those are the two answers. Hope this helps to familiarize you with the quadratic equation and
how it can be used to solve quadratic equations.
.