320(t)=5t^2

Solution for 320(t)=5t^2 equation:

320(t)=5t^2

We move all terms to the left:

320(t)-(5t^2)=0

determiningTheFunctionDomain
-5t^2+320t=0

a = -5; b = 320; c = 0;
Δ = b2-4ac
Δ = 3202-4·(-5)·0
Δ = 102400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{102400}=320$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(320)-320}{2*-5}=\frac{-640}{-10}
=+64
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(320)+320}{2*-5}=\frac{0}{-10}
=0
$