H=((-16t^2)+64t)+42

Solution for H=((-16t^2)+64t)+42 equation:

=((-16H^2)+64H)+42

We move all terms to the left:

-(((-16H^2)+64H)+42)=0


We calculate terms in parentheses: -(((-16H^2)+64H)+42), so:

((-16H^2)+64H)+42


We calculate terms in parentheses: +((-16H^2)+64H), so:

(-16H^2)+64H

We get rid of parentheses

-16H^2+64H

Back to the equation:

+(-16H^2+64H)


We get rid of parentheses

-16H^2+64H+42

Back to the equation:

-(-16H^2+64H+42)


We get rid of parentheses

16H^2-64H-42=0

a = 16; b = -64; c = -42;
Δ = b2-4ac
Δ = -642-4·16·(-42)
Δ = 6784
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:

$\sqrt{\Delta}=\sqrt{6784}=\sqrt{64*106}=\sqrt{64}*\sqrt{106}=8\sqrt{106}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-8\sqrt{106}}{2*16}=\frac{64-8\sqrt{106}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+8\sqrt{106}}{2*16}=\frac{64+8\sqrt{106}}{32}
$