3x2-20x+24=0

Solution for 3x2-20x+24=0 equation:

3x^2-20x+24=0

a = 3; b = -20; c = +24;
Δ = b2-4ac
Δ = -202-4·3·24
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{7}}{2*3}=\frac{20-4\sqrt{7}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{7}}{2*3}=\frac{20+4\sqrt{7}}{6}
$