H=3t+14t-5t^2

Solution for H=3t+14t-5t^2 equation:

=3H+14H-5H^2

We move all terms to the left:

-(3H+14H-5H^2)=0

We get rid of parentheses

5H^2-3H-14H=0

We add all the numbers together, and all the variables

5H^2-17H=0

a = 5; b = -17; c = 0;
Δ = b2-4ac
Δ = -172-4·5·0
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-17}{2*5}=\frac{0}{10}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+17}{2*5}=\frac{34}{10}
=3+2/5
$