H(3)=-16t^2+265

Solution for H(3)=-16t^2+265 equation:

(3)=-16H^2+265

We move all terms to the left:

(3)-(-16H^2+265)=0

We get rid of parentheses

16H^2-265+3=0

We add all the numbers together, and all the variables

16H^2-262=0

a = 16; b = 0; c = -262;
Δ = b2-4ac
Δ = 02-4·16·(-262)
Δ = 16768
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:

$\sqrt{\Delta}=\sqrt{16768}=\sqrt{64*262}=\sqrt{64}*\sqrt{262}=8\sqrt{262}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{262}}{2*16}=\frac{0-8\sqrt{262}}{32}
=-\frac{8\sqrt{262}}{32}
=-\frac{\sqrt{262}}{4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{262}}{2*16}=\frac{0+8\sqrt{262}}{32}
=\frac{8\sqrt{262}}{32}
=\frac{\sqrt{262}}{4}
$