3x^2+9x=24

Solution for 3x^2+9x=24 equation:

3x^2+9x=24

We move all terms to the left:

3x^2+9x-(24)=0

a = 3; b = 9; c = -24;
Δ = b2-4ac
Δ = 92-4·3·(-24)
Δ = 369
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:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{41}}{2*3}=\frac{-9-3\sqrt{41}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{41}}{2*3}=\frac{-9+3\sqrt{41}}{6}
$