-0.5x^2+60x-500=0

Solution for -0.5x^2+60x-500=0 equation:

-0.5x^2+60x-500=0

a = -0.5; b = 60; c = -500;
Δ = b2-4ac
Δ = 602-4·(-0.5)·(-500)
Δ = 2600
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:

$\sqrt{\Delta}=\sqrt{2600}=\sqrt{100*26}=\sqrt{100}*\sqrt{26}=10\sqrt{26}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-10\sqrt{26}}{2*-0.5}=\frac{-60-10\sqrt{26}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+10\sqrt{26}}{2*-0.5}=\frac{-60+10\sqrt{26}}{-1}
$