3x^2=20x+22

Solution for 3x^2=20x+22 equation:

3x^2=20x+22

We move all terms to the left:

3x^2-(20x+22)=0

We get rid of parentheses

3x^2-20x-22=0

a = 3; b = -20; c = -22;
Δ = b2-4ac
Δ = -202-4·3·(-22)
Δ = 664
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:

$\sqrt{\Delta}=\sqrt{664}=\sqrt{4*166}=\sqrt{4}*\sqrt{166}=2\sqrt{166}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{166}}{2*3}=\frac{20-2\sqrt{166}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{166}}{2*3}=\frac{20+2\sqrt{166}}{6}
$