H=5t2+14t+6

Solution for H=5t2+14t+6 equation:

=5H^2+14H+6

We move all terms to the left:

-(5H^2+14H+6)=0

We get rid of parentheses

-5H^2-14H-6=0

a = -5; b = -14; c = -6;
Δ = b2-4ac
Δ = -142-4·(-5)·(-6)
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{19}}{2*-5}=\frac{14-2\sqrt{19}}{-10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{19}}{2*-5}=\frac{14+2\sqrt{19}}{-10}
$