4=-16t^2+80t+224

Solution for 4=-16t^2+80t+224 equation:

4=-16t^2+80t+224

We move all terms to the left:

4-(-16t^2+80t+224)=0

We get rid of parentheses

16t^2-80t-224+4=0

We add all the numbers together, and all the variables

16t^2-80t-220=0

a = 16; b = -80; c = -220;
Δ = b2-4ac
Δ = -802-4·16·(-220)
Δ = 20480
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:

$\sqrt{\Delta}=\sqrt{20480}=\sqrt{4096*5}=\sqrt{4096}*\sqrt{5}=64\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-64\sqrt{5}}{2*16}=\frac{80-64\sqrt{5}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+64\sqrt{5}}{2*16}=\frac{80+64\sqrt{5}}{32}
$