H=-13t^2+10t+3

Solution for H=-13t^2+10t+3 equation:

=-13H^2+10H+3

We move all terms to the left:

-(-13H^2+10H+3)=0

We get rid of parentheses

13H^2-10H-3=0

a = 13; b = -10; c = -3;
Δ = b2-4ac
Δ = -102-4·13·(-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{-(-10)-16}{2*13}=\frac{-6}{26}
=-3/13
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+16}{2*13}=\frac{26}{26}
=1
$