2x^2-4x-8=10x-8

Solution for 2x^2-4x-8=10x-8 equation:

2x^2-4x-8=10x-8

We move all terms to the left:

2x^2-4x-8-(10x-8)=0

We get rid of parentheses

2x^2-4x-10x+8-8=0

We add all the numbers together, and all the variables

2x^2-14x=0

a = 2; b = -14; c = 0;
Δ = b2-4ac
Δ = -142-4·2·0
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-14}{2*2}=\frac{0}{4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+14}{2*2}=\frac{28}{4}
=7
$