3x^2-20x+34=8

Solution for 3x^2-20x+34=8 equation:

3x^2-20x+34=8

We move all terms to the left:

3x^2-20x+34-(8)=0

We add all the numbers together, and all the variables

3x^2-20x+26=0

a = 3; b = -20; c = +26;
Δ = b2-4ac
Δ = -202-4·3·26
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{22}}{2*3}=\frac{20-2\sqrt{22}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{22}}{2*3}=\frac{20+2\sqrt{22}}{6}
$