-200+10x-0.05x^2=0

Solution for -200+10x-0.05x^2=0 equation:

-200+10x-0.05x^2=0

a = -0.05; b = 10; c = -200;
Δ = b2-4ac
Δ = 102-4·(-0.05)·(-200)
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{15}}{2*-0.05}=\frac{-10-2\sqrt{15}}{-0.1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{15}}{2*-0.05}=\frac{-10+2\sqrt{15}}{-0.1}
$