H(12)=-16t^2+20

Solution for H(12)=-16t^2+20 equation:

(12)=-16H^2+20

We move all terms to the left:

(12)-(-16H^2+20)=0

We get rid of parentheses

16H^2-20+12=0

We add all the numbers together, and all the variables

16H^2-8=0

a = 16; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·16·(-8)
Δ = 512
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{2}}{2*16}=\frac{0-16\sqrt{2}}{32}
=-\frac{16\sqrt{2}}{32}
=-\frac{\sqrt{2}}{2}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{2}}{2*16}=\frac{0+16\sqrt{2}}{32}
=\frac{16\sqrt{2}}{32}
=\frac{\sqrt{2}}{2}
$