H=3+14t+-5t2

Solution for H=3+14t+-5t2 equation:

=3+14H+-5H^2

We move all terms to the left:

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

We use the square of the difference formula

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

We get rid of parentheses

5H^2-14H-3=0

a = 5; b = -14; c = -3;
Δ = b2-4ac
Δ = -142-4·5·(-3)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-16}{2*5}=\frac{-2}{10}
=-1/5
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+16}{2*5}=\frac{30}{10}
=3
$