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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3H^2-12H-10=0

a = 3; b = -12; c = -10;
Δ = b2-4ac
Δ = -122-4·3·(-10)
Δ = 264
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:

$\sqrt{\Delta}=\sqrt{264}=\sqrt{4*66}=\sqrt{4}*\sqrt{66}=2\sqrt{66}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{66}}{2*3}=\frac{12-2\sqrt{66}}{6}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{66}}{2*3}=\frac{12+2\sqrt{66}}{6}
$