H=1+7t-5t^2

Solution for H=1+7t-5t^2 equation:

=1+7H-5H^2

We move all terms to the left:

-(1+7H-5H^2)=0

We get rid of parentheses

5H^2-7H-1=0

a = 5; b = -7; c = -1;
Δ = b2-4ac
Δ = -72-4·5·(-1)
Δ = 69
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{69}}{2*5}=\frac{7-\sqrt{69}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{69}}{2*5}=\frac{7+\sqrt{69}}{10}
$