(t)=-4t^2+16+20

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

(t)=-4t^2+16+20

We move all terms to the left:

(t)-(-4t^2+16+20)=0

We get rid of parentheses

4t^2+t-16-20=0

We add all the numbers together, and all the variables

4t^2+t-36=0

a = 4; b = 1; c = -36;
Δ = b2-4ac
Δ = 12-4·4·(-36)
Δ = 577
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{577}}{2*4}=\frac{-1-\sqrt{577}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{577}}{2*4}=\frac{-1+\sqrt{577}}{8}
$