H(1)=-16(x)^2+68x+4

Solution for H(1)=-16(x)^2+68x+4 equation:

(1)=-16(H)^2+68H+4

We move all terms to the left:

(1)-(-16(H)^2+68H+4)=0

We get rid of parentheses

16H^2-68H-4+1=0

We add all the numbers together, and all the variables

16H^2-68H-3=0

a = 16; b = -68; c = -3;
Δ = b2-4ac
Δ = -682-4·16·(-3)
Δ = 4816
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:

$\sqrt{\Delta}=\sqrt{4816}=\sqrt{16*301}=\sqrt{16}*\sqrt{301}=4\sqrt{301}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-68)-4\sqrt{301}}{2*16}=\frac{68-4\sqrt{301}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-68)+4\sqrt{301}}{2*16}=\frac{68+4\sqrt{301}}{32}
$