x^2-19x-8=-8x+4

Solution for x^2-19x-8=-8x+4 equation:

x^2-19x-8=-8x+4

We move all terms to the left:

x^2-19x-8-(-8x+4)=0

We get rid of parentheses

x^2-19x+8x-4-8=0

We add all the numbers together, and all the variables

x^2-11x-12=0

a = 1; b = -11; c = -12;
Δ = b2-4ac
Δ = -112-4·1·(-12)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-13}{2*1}=\frac{-2}{2}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+13}{2*1}=\frac{24}{2}
=12
$