H(t)=-4t^2+16tH(2)=

Solution for H(t)=-4t^2+16tH(2)= equation:

(H)=-4H^2+16H(2)=

We move all terms to the left:

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

We get rid of parentheses

4H^2-16H2+H=0

We add all the numbers together, and all the variables

-12H^2+H=0

a = -12; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-12)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-12}=\frac{-2}{-24}
=1/12
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-12}=\frac{0}{-24}
=0
$