3x^2+24x=-15

Solution for 3x^2+24x=-15 equation:

3x^2+24x=-15

We move all terms to the left:

3x^2+24x-(-15)=0

We add all the numbers together, and all the variables

3x^2+24x+15=0

a = 3; b = 24; c = +15;
Δ = b2-4ac
Δ = 242-4·3·15
Δ = 396
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:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-6\sqrt{11}}{2*3}=\frac{-24-6\sqrt{11}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+6\sqrt{11}}{2*3}=\frac{-24+6\sqrt{11}}{6}
$