Posted by **Efrain** on Sunday, February 17, 2013 at 4:18pm.

This is my question 2x^2-8x=-25

I cant seem to figure out how to solve for x?

- College Algebra -
**Unkown**, Sunday, February 17, 2013 at 4:37pm
Add the -25 to the other side...

2x^2-8x+25=0,

Now use the quadratic formula to solve for x...

Hope this helps

- College Algebra -
**Efrain**, Sunday, February 17, 2013 at 7:00pm
I was told to start with

x^2-8x+16=7/2 so where do i go from there?

- College Algebra -
**Reiny**, Sunday, February 17, 2013 at 7:12pm
2x^2 - 8x =-25

divide each term by 2

using completing the square,

x^2 - 4x = -25/2

take 1/2 the coefficient of the x term, square it, then add it to each side

x^2 - 4x **+4 ** = -25/2 **+4 **

(x-2)^2 = -17/2 or -34/4 , (I anticipated to take √ of both sides)

x -2 = ± √-34/2 = ± i√34/2

x = 4/2 ± i√34/2 = (4 ± i√34)/2

or ... by the formula

2x^2 - 8x + 25 = 0

x = (8 ± √-136)/4

= (8 ± 2√-34)/2

= (4 ± i√34)/2

- College Algebra -
**Efrain**, Sunday, February 17, 2013 at 7:54pm
I was told to start with

x^2-8x+16=7/2 so where do i go from there? That's after / by 2.

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