3x^2+20x-64=0

Solution for 3x^2+20x-64=0 equation:

3x^2+20x-64=0

a = 3; b = 20; c = -64;
Δ = b2-4ac
Δ = 202-4·3·(-64)
Δ = 1168
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:

$\sqrt{\Delta}=\sqrt{1168}=\sqrt{16*73}=\sqrt{16}*\sqrt{73}=4\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{73}}{2*3}=\frac{-20-4\sqrt{73}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{73}}{2*3}=\frac{-20+4\sqrt{73}}{6}
$