H=2t^2+10t+2

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

=2H^2+10H+2

We move all terms to the left:

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

We get rid of parentheses

-2H^2-10H-2=0

a = -2; b = -10; c = -2;
Δ = b2-4ac
Δ = -102-4·(-2)·(-2)
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{21}}{2*-2}=\frac{10-2\sqrt{21}}{-4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{21}}{2*-2}=\frac{10+2\sqrt{21}}{-4}
$