H(t)=-25t^2-5t+1269

Solution for H(t)=-25t^2-5t+1269 equation:

(H)=-25H^2-5H+1269

We move all terms to the left:

(H)-(-25H^2-5H+1269)=0

We get rid of parentheses

25H^2+5H+H-1269=0

We add all the numbers together, and all the variables

25H^2+6H-1269=0

a = 25; b = 6; c = -1269;
Δ = b2-4ac
Δ = 62-4·25·(-1269)
Δ = 126936
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:

$\sqrt{\Delta}=\sqrt{126936}=\sqrt{36*3526}=\sqrt{36}*\sqrt{3526}=6\sqrt{3526}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{3526}}{2*25}=\frac{-6-6\sqrt{3526}}{50}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{3526}}{2*25}=\frac{-6+6\sqrt{3526}}{50}
$