H(t)=5t^2+24t+3/2

Solution for H(t)=5t^2+24t+3/2 equation:

(H)=5H^2+24H+3/2

We move all terms to the left:

(H)-(5H^2+24H+3/2)=0

We get rid of parentheses

-5H^2+H-24H-3/2=0

We multiply all the terms by the denominator


-5H^2*2+H*2-24H*2-3=0

Wy multiply elements

-10H^2+2H-48H-3=0

We add all the numbers together, and all the variables

-10H^2-46H-3=0

a = -10; b = -46; c = -3;
Δ = b2-4ac
Δ = -462-4·(-10)·(-3)
Δ = 1996
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:

$\sqrt{\Delta}=\sqrt{1996}=\sqrt{4*499}=\sqrt{4}*\sqrt{499}=2\sqrt{499}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-46)-2\sqrt{499}}{2*-10}=\frac{46-2\sqrt{499}}{-20}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-46)+2\sqrt{499}}{2*-10}=\frac{46+2\sqrt{499}}{-20}
$