-36=3x^2+21x

Solution for -36=3x^2+21x equation:

-36=3x^2+21x

We move all terms to the left:

-36-(3x^2+21x)=0

We get rid of parentheses

-3x^2-21x-36=0

a = -3; b = -21; c = -36;
Δ = b2-4ac
Δ = -212-4·(-3)·(-36)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-3}{2*-3}=\frac{18}{-6}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+3}{2*-3}=\frac{24}{-6}
=-4
$