2500+10x-0.2x^2=0

Solution for 2500+10x-0.2x^2=0 equation:

2500+10x-0.2x^2=0

a = -0.2; b = 10; c = +2500;
Δ = b2-4ac
Δ = 102-4·(-0.2)·2500
Δ = 2100
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:

$\sqrt{\Delta}=\sqrt{2100}=\sqrt{100*21}=\sqrt{100}*\sqrt{21}=10\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{21}}{2*-0.2}=\frac{-10-10\sqrt{21}}{-0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{21}}{2*-0.2}=\frac{-10+10\sqrt{21}}{-0.4}
$