H(3)=-16x^2+64x+80

Solution for H(3)=-16x^2+64x+80 equation:

(3)=-16H^2+64H+80

We move all terms to the left:

(3)-(-16H^2+64H+80)=0

We get rid of parentheses

16H^2-64H-80+3=0

We add all the numbers together, and all the variables

16H^2-64H-77=0

a = 16; b = -64; c = -77;
Δ = b2-4ac
Δ = -642-4·16·(-77)
Δ = 9024
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:

$\sqrt{\Delta}=\sqrt{9024}=\sqrt{64*141}=\sqrt{64}*\sqrt{141}=8\sqrt{141}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-8\sqrt{141}}{2*16}=\frac{64-8\sqrt{141}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+8\sqrt{141}}{2*16}=\frac{64+8\sqrt{141}}{32}
$