t^2+25t-31.25=0

Solution for t^2+25t-31.25=0 equation:

t^2+25t-31.25=0

a = 1; b = 25; c = -31.25;
Δ = b2-4ac
Δ = 252-4·1·(-31.25)
Δ = 750
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:

$\sqrt{\Delta}=\sqrt{750}=\sqrt{25*30}=\sqrt{25}*\sqrt{30}=5\sqrt{30}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-5\sqrt{30}}{2*1}=\frac{-25-5\sqrt{30}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+5\sqrt{30}}{2*1}=\frac{-25+5\sqrt{30}}{2}
$