-.2b^2+15b-250=0

Solution for -.2b^2+15b-250=0 equation:

-.2b^2+15b-250=0

We add all the numbers together, and all the variables

-0.2b^2+15b-250=0

a = -0.2; b = 15; c = -250;
Δ = b2-4ac
Δ = 152-4·(-0.2)·(-250)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-5}{2*-0.2}=\frac{-20}{-0.4}
=+50
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+5}{2*-0.2}=\frac{-10}{-0.4}
=+25
$