t^2-850t+2500=0

Solution for t^2-850t+2500=0 equation:

t^2-850t+2500=0

a = 1; b = -850; c = +2500;
Δ = b2-4ac
Δ = -8502-4·1·2500
Δ = 712500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{712500}=\sqrt{2500*285}=\sqrt{2500}*\sqrt{285}=50\sqrt{285}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-850)-50\sqrt{285}}{2*1}=\frac{850-50\sqrt{285}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-850)+50\sqrt{285}}{2*1}=\frac{850+50\sqrt{285}}{2}
$