-0.01x^2+25x-1525=0

Solution for -0.01x^2+25x-1525=0 equation:

-0.01x^2+25x-1525=0

a = -0.01; b = 25; c = -1525;
Δ = b2-4ac
Δ = 252-4·(-0.01)·(-1525)
Δ = 564
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:

$\sqrt{\Delta}=\sqrt{564}=\sqrt{4*141}=\sqrt{4}*\sqrt{141}=2\sqrt{141}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-2\sqrt{141}}{2*-0.01}=\frac{-25-2\sqrt{141}}{-0.02}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+2\sqrt{141}}{2*-0.01}=\frac{-25+2\sqrt{141}}{-0.02}
$