3x2+18x=24

Solution for 3x2+18x=24 equation:

3x^2+18x=24

We move all terms to the left:

3x^2+18x-(24)=0

a = 3; b = 18; c = -24;
Δ = b2-4ac
Δ = 182-4·3·(-24)
Δ = 612
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:

$\sqrt{\Delta}=\sqrt{612}=\sqrt{36*17}=\sqrt{36}*\sqrt{17}=6\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{17}}{2*3}=\frac{-18-6\sqrt{17}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{17}}{2*3}=\frac{-18+6\sqrt{17}}{6}
$