(4t^2)+80t+76=0

Solution for (4t^2)+80t+76=0 equation:

(4t^2)+80t+76=0

a = 4; b = 80; c = +76;
Δ = b2-4ac
Δ = 802-4·4·76
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-72}{2*4}=\frac{-152}{8}
=-19
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+72}{2*4}=\frac{-8}{8}
=-1
$