-10b^2+33b-20=0

Solution for -10b^2+33b-20=0 equation:

-10b^2+33b-20=0

a = -10; b = 33; c = -20;
Δ = b2-4ac
Δ = 332-4·(-10)·(-20)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-17}{2*-10}=\frac{-50}{-20}
=2+1/2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+17}{2*-10}=\frac{-16}{-20}
=4/5
$