t^2+4t-36=0

Solution for t^2+4t-36=0 equation:

t^2+4t-36=0

a = 1; b = 4; c = -36;
Δ = b2-4ac
Δ = 42-4·1·(-36)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{10}}{2*1}=\frac{-4-4\sqrt{10}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{10}}{2*1}=\frac{-4+4\sqrt{10}}{2}
$