t^2+20t-20=0

Solution for t^2+20t-20=0 equation:

t^2+20t-20=0

a = 1; b = 20; c = -20;
Δ = b2-4ac
Δ = 202-4·1·(-20)
Δ = 480
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:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{30}}{2*1}=\frac{-20-4\sqrt{30}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{30}}{2*1}=\frac{-20+4\sqrt{30}}{2}
$