-20b^2-9b+18=0

Solution for -20b^2-9b+18=0 equation:

-20b^2-9b+18=0

a = -20; b = -9; c = +18;
Δ = b2-4ac
Δ = -92-4·(-20)·18
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-39}{2*-20}=\frac{-30}{-40}
=3/4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+39}{2*-20}=\frac{48}{-40}
=-1+1/5
$