SOLUTION: Solve 2x^2-5x=3 Having trouble..any help appreciated.

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Question 88032: Solve
2x^2-5x=3
Having trouble..any help appreciated.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
2+x+%5E+2+-+5+x+=+3+ Start with the given equation


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

Now let's use the quadratic formula to solve for x:
Starting with the general quadratic

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

the general form of 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+=+%285+%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+25-4%2A2%2A-3+%29%29%2F%282%2A2%29 Square -5 to get 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



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=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=-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

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

+graph%28+500%2C+500%2C+-11%2C+13%2C+-11%2C+13%2C2%2Ax%5E2%2B-5%2Ax%2B-3%29+

and we can see that the roots are x=3 and x=-1%2F2. This verifies our answer