H(t)=-16t^2+11t+4

Solution for H(t)=-16t^2+11t+4 equation:

(H)=-16H^2+11H+4

We move all terms to the left:

(H)-(-16H^2+11H+4)=0

We get rid of parentheses

16H^2-11H+H-4=0

We add all the numbers together, and all the variables

16H^2-10H-4=0

a = 16; b = -10; c = -4;
Δ = b2-4ac
Δ = -102-4·16·(-4)
Δ = 356
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:

$\sqrt{\Delta}=\sqrt{356}=\sqrt{4*89}=\sqrt{4}*\sqrt{89}=2\sqrt{89}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{89}}{2*16}=\frac{10-2\sqrt{89}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{89}}{2*16}=\frac{10+2\sqrt{89}}{32}
$