2x^2-x=-9x+8

Solution for 2x^2-x=-9x+8 equation:

2x^2-x=-9x+8

We move all terms to the left:

2x^2-x-(-9x+8)=0

We add all the numbers together, and all the variables

2x^2-1x-(-9x+8)=0

We get rid of parentheses

2x^2-1x+9x-8=0

We add all the numbers together, and all the variables

2x^2+8x-8=0

a = 2; b = 8; c = -8;
Δ = b2-4ac
Δ = 82-4·2·(-8)
Δ = 128
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{2}}{2*2}=\frac{-8-8\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{2}}{2*2}=\frac{-8+8\sqrt{2}}{4}
$