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

Solution for 3x^2+(-24x)=0 equation:

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

We get rid of parentheses

3x^2-24x=0

a = 3; b = -24; c = 0;
Δ = b2-4ac
Δ = -242-4·3·0
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-24}{2*3}=\frac{0}{6}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+24}{2*3}=\frac{48}{6}
=8
$