t^2+25t-300=0

Solution for t^2+25t-300=0 equation:

t^2+25t-300=0

a = 1; b = 25; c = -300;
Δ = b2-4ac
Δ = 252-4·1·(-300)
Δ = 1825
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:

$\sqrt{\Delta}=\sqrt{1825}=\sqrt{25*73}=\sqrt{25}*\sqrt{73}=5\sqrt{73}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-5\sqrt{73}}{2*1}=\frac{-25-5\sqrt{73}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+5\sqrt{73}}{2*1}=\frac{-25+5\sqrt{73}}{2}
$