20=3x^2+6x

Solution for 20=3x^2+6x equation:

20=3x^2+6x

We move all terms to the left:

20-(3x^2+6x)=0

We get rid of parentheses

-3x^2-6x+20=0

a = -3; b = -6; c = +20;
Δ = b2-4ac
Δ = -62-4·(-3)·20
Δ = 276
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:

$\sqrt{\Delta}=\sqrt{276}=\sqrt{4*69}=\sqrt{4}*\sqrt{69}=2\sqrt{69}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{69}}{2*-3}=\frac{6-2\sqrt{69}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{69}}{2*-3}=\frac{6+2\sqrt{69}}{-6}
$