225(t)=16t^2+225

Solution for 225(t)=16t^2+225 equation:

225(t)=16t^2+225

We move all terms to the left:

225(t)-(16t^2+225)=0

We get rid of parentheses

-16t^2+225t-225=0

a = -16; b = 225; c = -225;
Δ = b2-4ac
Δ = 2252-4·(-16)·(-225)
Δ = 36225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{36225}=\sqrt{225*161}=\sqrt{225}*\sqrt{161}=15\sqrt{161}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(225)-15\sqrt{161}}{2*-16}=\frac{-225-15\sqrt{161}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(225)+15\sqrt{161}}{2*-16}=\frac{-225+15\sqrt{161}}{-32}
$