20=-4t^2+16t+7

Solution for 20=-4t^2+16t+7 equation:

20=-4t^2+16t+7

We move all terms to the left:

20-(-4t^2+16t+7)=0

We get rid of parentheses

4t^2-16t-7+20=0

We add all the numbers together, and all the variables

4t^2-16t+13=0

a = 4; b = -16; c = +13;
Δ = b2-4ac
Δ = -162-4·4·13
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{3}}{2*4}=\frac{16-4\sqrt{3}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{3}}{2*4}=\frac{16+4\sqrt{3}}{8}
$