-3b2+3b+90=0

Solution for -3b2+3b+90=0 equation:

-3b^2+3b+90=0

a = -3; b = 3; c = +90;
Δ = b2-4ac
Δ = 32-4·(-3)·90
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-33}{2*-3}=\frac{-36}{-6}
=+6
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+33}{2*-3}=\frac{30}{-6}
=-5
$