t^2-25t-125=0

Solution for t^2-25t-125=0 equation:

t^2-25t-125=0

a = 1; b = -25; c = -125;
Δ = b2-4ac
Δ = -252-4·1·(-125)
Δ = 1125
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:

$\sqrt{\Delta}=\sqrt{1125}=\sqrt{225*5}=\sqrt{225}*\sqrt{5}=15\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-15\sqrt{5}}{2*1}=\frac{25-15\sqrt{5}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+15\sqrt{5}}{2*1}=\frac{25+15\sqrt{5}}{2}
$