(3x^2)+24=240

Solution for (3x^2)+24=240 equation:

(3x^2)+24=240

We move all terms to the left:

(3x^2)+24-(240)=0

We add all the numbers together, and all the variables

3x^2-216=0

a = 3; b = 0; c = -216;
Δ = b2-4ac
Δ = 02-4·3·(-216)
Δ = 2592
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:

$\sqrt{\Delta}=\sqrt{2592}=\sqrt{1296*2}=\sqrt{1296}*\sqrt{2}=36\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-36\sqrt{2}}{2*3}=\frac{0-36\sqrt{2}}{6}
=-\frac{36\sqrt{2}}{6}
=-6\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+36\sqrt{2}}{2*3}=\frac{0+36\sqrt{2}}{6}
=\frac{36\sqrt{2}}{6}
=6\sqrt{2}
$