H=3(4t-2)(-t+4)

Solution for H=3(4t-2)(-t+4) equation:

=3(4H-2)(-H+4)

We move all terms to the left:

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

We add all the numbers together, and all the variables

-(3(4H-2)(-1H+4))=0

We multiply parentheses ..

-(3(-4H^2+16H+2H-8))=0


We calculate terms in parentheses: -(3(-4H^2+16H+2H-8)), so:

3(-4H^2+16H+2H-8)

We multiply parentheses

-12H^2+48H+6H-24

We add all the numbers together, and all the variables

-12H^2+54H-24

Back to the equation:

-(-12H^2+54H-24)


We get rid of parentheses

12H^2-54H+24=0

a = 12; b = -54; c = +24;
Δ = b2-4ac
Δ = -542-4·12·24
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-54)-42}{2*12}=\frac{12}{24}
=1/2
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-54)+42}{2*12}=\frac{96}{24}
=4
$