-0,03x^2+40x-150=0

Solution for -0,03x^2+40x-150=0 equation:

-0.03x^2+40x-150=0

a = -0.03; b = 40; c = -150;
Δ = b2-4ac
Δ = 402-4·(-0.03)·(-150)
Δ = 1582
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-\sqrt{1582}}{2*-0.03}=\frac{-40-\sqrt{1582}}{-0.06}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+\sqrt{1582}}{2*-0.03}=\frac{-40+\sqrt{1582}}{-0.06}
$