H(t)=-(4t-12)(4t-36)

Solution for H(t)=-(4t-12)(4t-36) equation:

(H)=-(4H-12)(4H-36)

We move all terms to the left:

(H)-(-(4H-12)(4H-36))=0

We multiply parentheses ..

-(-(+16H^2-144H-48H+432))+H=0


We calculate terms in parentheses: -(-(+16H^2-144H-48H+432)), so:

-(+16H^2-144H-48H+432)

We get rid of parentheses

-16H^2+144H+48H-432

We add all the numbers together, and all the variables

-16H^2+192H-432

Back to the equation:

-(-16H^2+192H-432)


We get rid of parentheses

16H^2-192H+H+432=0

We add all the numbers together, and all the variables

16H^2-191H+432=0

a = 16; b = -191; c = +432;
Δ = b2-4ac
Δ = -1912-4·16·432
Δ = 8833
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:

$\sqrt{\Delta}=\sqrt{8833}=\sqrt{121*73}=\sqrt{121}*\sqrt{73}=11\sqrt{73}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-191)-11\sqrt{73}}{2*16}=\frac{191-11\sqrt{73}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-191)+11\sqrt{73}}{2*16}=\frac{191+11\sqrt{73}}{32}
$