t^2+5t-25=0

Solution for t^2+5t-25=0 equation:

t^2+5t-25=0

a = 1; b = 5; c = -25;
Δ = b2-4ac
Δ = 52-4·1·(-25)
Δ = 125
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:

$\sqrt{\Delta}=\sqrt{125}=\sqrt{25*5}=\sqrt{25}*\sqrt{5}=5\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{5}}{2*1}=\frac{-5-5\sqrt{5}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{5}}{2*1}=\frac{-5+5\sqrt{5}}{2}
$