H(t)=-5t^2+50t+6

Solution for H(t)=-5t^2+50t+6 equation:

(H)=-5H^2+50H+6

We move all terms to the left:

(H)-(-5H^2+50H+6)=0

We get rid of parentheses

5H^2-50H+H-6=0

We add all the numbers together, and all the variables

5H^2-49H-6=0

a = 5; b = -49; c = -6;
Δ = b2-4ac
Δ = -492-4·5·(-6)
Δ = 2521
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{-(-49)-\sqrt{2521}}{2*5}=\frac{49-\sqrt{2521}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+\sqrt{2521}}{2*5}=\frac{49+\sqrt{2521}}{10}
$