H=-2t(8t-22)+0

Solution for H=-2t(8t-22)+0 equation:

=-2H(8H-22)+0

We move all terms to the left:

-(-2H(8H-22)+0)=0


We calculate terms in parentheses: -(-2H(8H-22)+0), so:

-2H(8H-22)+0

We add all the numbers together, and all the variables

-2H(8H-22)

We multiply parentheses

-16H^2+44H

Back to the equation:

-(-16H^2+44H)


We get rid of parentheses

16H^2-44H=0

a = 16; b = -44; c = 0;
Δ = b2-4ac
Δ = -442-4·16·0
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-44)-44}{2*16}=\frac{0}{32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-44)+44}{2*16}=\frac{88}{32}
=2+3/4
$