College Algebra

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This is my question 2x^2-8x=-25
I cant seem to figure out how to solve for x?

• College Algebra -

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 -

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

• College Algebra -

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 -

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

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