-3b^2+15b+18=0

Solution for -3b^2+15b+18=0 equation:

-3b^2+15b+18=0

a = -3; b = 15; c = +18;
Δ = b2-4ac
Δ = 152-4·(-3)·18
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-21}{2*-3}=\frac{-36}{-6}
=+6
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+21}{2*-3}=\frac{6}{-6}
=-1
$