36=3x^2+28x+24

Solution for 36=3x^2+28x+24 equation:

36=3x^2+28x+24

We move all terms to the left:

36-(3x^2+28x+24)=0

We get rid of parentheses

-3x^2-28x-24+36=0

We add all the numbers together, and all the variables

-3x^2-28x+12=0

a = -3; b = -28; c = +12;
Δ = b2-4ac
Δ = -282-4·(-3)·12
Δ = 928
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:

$\sqrt{\Delta}=\sqrt{928}=\sqrt{16*58}=\sqrt{16}*\sqrt{58}=4\sqrt{58}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{58}}{2*-3}=\frac{28-4\sqrt{58}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{58}}{2*-3}=\frac{28+4\sqrt{58}}{-6}
$