x^2-10x+12=68

Solution for x^2-10x+12=68 equation:

x^2-10x+12=68

We move all terms to the left:

x^2-10x+12-(68)=0

We add all the numbers together, and all the variables

x^2-10x-56=0

a = 1; b = -10; c = -56;
Δ = b2-4ac
Δ = -102-4·1·(-56)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-18}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+18}{2*1}=\frac{28}{2}
=14
$