-3x^2+4x+20=0

Solution for -3x^2+4x+20=0 equation:

-3x^2+4x+20=0

a = -3; b = 4; c = +20;
Δ = b2-4ac
Δ = 42-4·(-3)·20
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-16}{2*-3}=\frac{-20}{-6}
=3+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+16}{2*-3}=\frac{12}{-6}
=-2
$