H(t)=-4t2+18t+4

Solution for H(t)=-4t2+18t+4 equation:

(H)=-4H^2+18H+4

We move all terms to the left:

(H)-(-4H^2+18H+4)=0

We get rid of parentheses

4H^2-18H+H-4=0

We add all the numbers together, and all the variables

4H^2-17H-4=0

a = 4; b = -17; c = -4;
Δ = b2-4ac
Δ = -172-4·4·(-4)
Δ = 353
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{-(-17)-\sqrt{353}}{2*4}=\frac{17-\sqrt{353}}{8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{353}}{2*4}=\frac{17+\sqrt{353}}{8}
$