4t^2+20t-16=0

Solution for 4t^2+20t-16=0 equation:

4t^2+20t-16=0

a = 4; b = 20; c = -16;
Δ = b2-4ac
Δ = 202-4·4·(-16)
Δ = 656
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:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{41}}{2*4}=\frac{-20-4\sqrt{41}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{41}}{2*4}=\frac{-20+4\sqrt{41}}{8}
$