H(t)=-4t^2+16t+48

Solution for H(t)=-4t^2+16t+48 equation:

(H)=-4H^2+16H+48

We move all terms to the left:

(H)-(-4H^2+16H+48)=0

We get rid of parentheses

4H^2-16H+H-48=0

We add all the numbers together, and all the variables

4H^2-15H-48=0

a = 4; b = -15; c = -48;
Δ = b2-4ac
Δ = -152-4·4·(-48)
Δ = 993
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{993}}{2*4}=\frac{15-\sqrt{993}}{8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{993}}{2*4}=\frac{15+\sqrt{993}}{8}
$