32t^2=4t

Solution for 32t^2=4t equation:

32t^2=4t

We move all terms to the left:

32t^2-(4t)=0

a = 32; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·32·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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