H(t)=4t2+48t+3

Solution for H(t)=4t2+48t+3 equation:

(H)=4H^2+48H+3

We move all terms to the left:

(H)-(4H^2+48H+3)=0

We get rid of parentheses

-4H^2+H-48H-3=0

We add all the numbers together, and all the variables

-4H^2-47H-3=0

a = -4; b = -47; c = -3;
Δ = b2-4ac
Δ = -472-4·(-4)·(-3)
Δ = 2161
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{-(-47)-\sqrt{2161}}{2*-4}=\frac{47-\sqrt{2161}}{-8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+\sqrt{2161}}{2*-4}=\frac{47+\sqrt{2161}}{-8}
$