H=-2t^2+16t-14

Solution for H=-2t^2+16t-14 equation:

=-2H^2+16H-14

We move all terms to the left:

-(-2H^2+16H-14)=0

We get rid of parentheses

2H^2-16H+14=0

a = 2; b = -16; c = +14;
Δ = b2-4ac
Δ = -162-4·2·14
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-12}{2*2}=\frac{4}{4}
=1
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+12}{2*2}=\frac{28}{4}
=7
$