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

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

=-2H^2+10H+3

We move all terms to the left:

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

We get rid of parentheses

2H^2-10H-3=0

a = 2; b = -10; c = -3;
Δ = b2-4ac
Δ = -102-4·2·(-3)
Δ = 124
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{124}=\sqrt{4*31}=\sqrt{4}*\sqrt{31}=2\sqrt{31}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{31}}{2*2}=\frac{10-2\sqrt{31}}{4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{31}}{2*2}=\frac{10+2\sqrt{31}}{4}
$