-0.2b^2+15b-150=0

Solution for -0.2b^2+15b-150=0 equation:

-0.2b^2+15b-150=0

a = -0.2; b = 15; c = -150;
Δ = b2-4ac
Δ = 152-4·(-0.2)·(-150)
Δ = 105
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{105}}{2*-0.2}=\frac{-15-\sqrt{105}}{-0.4}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{105}}{2*-0.2}=\frac{-15+\sqrt{105}}{-0.4}
$