H=-t2+10t+24

Solution for H=-t2+10t+24 equation:

=-H2+10H+24

We move all terms to the left:

-(-H2+10H+24)=0

We add all the numbers together, and all the variables

-(-1H^2+10H+24)=0

We get rid of parentheses

1H^2-10H-24=0

We add all the numbers together, and all the variables

H^2-10H-24=0

a = 1; b = -10; c = -24;
Δ = b2-4ac
Δ = -102-4·1·(-24)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-14}{2*1}=\frac{-4}{2}
=-2
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+14}{2*1}=\frac{24}{2}
=12
$