H(t)=-t2+16t+63

Solution for H(t)=-t2+16t+63 equation:

(H)=-H2+16H+63

We move all terms to the left:

(H)-(-H2+16H+63)=0

We add all the numbers together, and all the variables

-(-1H^2+16H+63)+H=0

We get rid of parentheses

1H^2-16H+H-63=0

We add all the numbers together, and all the variables

H^2-15H-63=0

a = 1; b = -15; c = -63;
Δ = b2-4ac
Δ = -152-4·1·(-63)
Δ = 477
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:

$\sqrt{\Delta}=\sqrt{477}=\sqrt{9*53}=\sqrt{9}*\sqrt{53}=3\sqrt{53}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3\sqrt{53}}{2*1}=\frac{15-3\sqrt{53}}{2}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3\sqrt{53}}{2*1}=\frac{15+3\sqrt{53}}{2}
$