H=16t^2-4t-70

Solution for H=16t^2-4t-70 equation:

=16H^2-4H-70

We move all terms to the left:

-(16H^2-4H-70)=0

We get rid of parentheses

-16H^2+4H+70=0

a = -16; b = 4; c = +70;
Δ = b2-4ac
Δ = 42-4·(-16)·70
Δ = 4496
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:

$\sqrt{\Delta}=\sqrt{4496}=\sqrt{16*281}=\sqrt{16}*\sqrt{281}=4\sqrt{281}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{281}}{2*-16}=\frac{-4-4\sqrt{281}}{-32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{281}}{2*-16}=\frac{-4+4\sqrt{281}}{-32}
$