t^2+16t-300=0

Solution for t^2+16t-300=0 equation:

t^2+16t-300=0

a = 1; b = 16; c = -300;
Δ = b2-4ac
Δ = 162-4·1·(-300)
Δ = 1456
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:

$\sqrt{\Delta}=\sqrt{1456}=\sqrt{16*91}=\sqrt{16}*\sqrt{91}=4\sqrt{91}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{91}}{2*1}=\frac{-16-4\sqrt{91}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{91}}{2*1}=\frac{-16+4\sqrt{91}}{2}
$