H=-4(t-3)(4t+5)

Solution for H=-4(t-3)(4t+5) equation:

=-4(H-3)(4H+5)

We move all terms to the left:

-(-4(H-3)(4H+5))=0

We multiply parentheses ..

-(-4(+4H^2+5H-12H-15))=0


We calculate terms in parentheses: -(-4(+4H^2+5H-12H-15)), so:

-4(+4H^2+5H-12H-15)

We multiply parentheses

-16H^2-20H+48H+60

We add all the numbers together, and all the variables

-16H^2+28H+60

Back to the equation:

-(-16H^2+28H+60)


We get rid of parentheses

16H^2-28H-60=0

a = 16; b = -28; c = -60;
Δ = b2-4ac
Δ = -282-4·16·(-60)
Δ = 4624
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4624}=68$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-68}{2*16}=\frac{-40}{32}
=-1+1/4
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+68}{2*16}=\frac{96}{32}
=3
$