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 = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-12}{2*4}=\frac{-32}{8}
=-4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+12}{2*4}=\frac{-8}{8}
=-1
$