-0.025x^2+12x-1000=0

Solution for -0.025x^2+12x-1000=0 equation:

-0.025x^2+12x-1000=0

a = -0.025; b = 12; c = -1000;
Δ = b2-4ac
Δ = 122-4·(-0.025)·(-1000)
Δ = 44
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{11}}{2*-0.025}=\frac{-12-2\sqrt{11}}{-0.05}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{11}}{2*-0.025}=\frac{-12+2\sqrt{11}}{-0.05}
$