SOLUTION: 2x^2-5x=3 I am having trouble figuring this out.I am suppose to use the quadratic formula

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Question 108812: 2x^2-5x=3 I am having trouble figuring this out.I am suppose to use the quadratic formula
Found 2 solutions by jim_thompson5910, MathLover1:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
2x%5E2-5x=3 Start with the given equation

2x%5E2-5x-3=0 Subtract 3 from both sides


Solved by pluggable solver: Quadratic Formula
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 2%2Ax%5E2-5%2Ax-3=0 ( notice a=2, b=-5, and c=-3)





x+=+%28--5+%2B-+sqrt%28+%28-5%29%5E2-4%2A2%2A-3+%29%29%2F%282%2A2%29 Plug in a=2, b=-5, and c=-3




x+=+%285+%2B-+sqrt%28+%28-5%29%5E2-4%2A2%2A-3+%29%29%2F%282%2A2%29 Negate -5 to get 5




x+=+%285+%2B-+sqrt%28+25-4%2A2%2A-3+%29%29%2F%282%2A2%29 Square -5 to get 25 (note: remember when you square -5, you must square the negative as well. This is because %28-5%29%5E2=-5%2A-5=25.)




x+=+%285+%2B-+sqrt%28+25%2B24+%29%29%2F%282%2A2%29 Multiply -4%2A-3%2A2 to get 24




x+=+%285+%2B-+sqrt%28+49+%29%29%2F%282%2A2%29 Combine like terms in the radicand (everything under the square root)




x+=+%285+%2B-+7%29%2F%282%2A2%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




x+=+%285+%2B-+7%29%2F4 Multiply 2 and 2 to get 4


So now the expression breaks down into two parts


x+=+%285+%2B+7%29%2F4 or x+=+%285+-+7%29%2F4


Lets look at the first part:


x=%285+%2B+7%29%2F4


x=12%2F4 Add the terms in the numerator

x=3 Divide


So one answer is

x=3




Now lets look at the second part:


x=%285+-+7%29%2F4


x=-2%2F4 Subtract the terms in the numerator

x=-1%2F2 Divide


So another answer is

x=-1%2F2


So our solutions are:

x=3 or x=-1%2F2


Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
2x%5E2+-5x+=+3
In the standard form ax%5E2+%2Bbx+%2Bc+=+0, your equation is:
+2x%5E2+-+5x+-3++=+0
=> you have coefficients:
=> a+=+2
=> b+=+-5
=> c+=+-3
Then roots are:
x%5B1%2C2%5D+=%28-b+%2B-+sqrt+%28b%5E2+-4%2Aa%2Ac+%29%29+%2F+%282%2Aa%29

x%5B1%2C2%5D+=%285+%2B-+sqrt+%2825+%2B+24%29%29+%2F+4
x%5B1%2C2%5D+=%285+%2B-+sqrt+%2849%29%29+%2F+4……………..sqrt%2849%29+=+7
x%5B1%2C2%5D+=%285+%2B-+7%29+%2F+4

x%5B1%5D+=%285+%2B+7%29+%2F+4
x%5B1%5D+=12+%2F+4
x%5B1%5D+=+3

x%5B2%5D+=%285+-+7%29+%2F+4
x%5B2%5D+=+-+2+%2F+4
x%5B2%5D+=+-+1+%2F+2