16t^2=20t

Solution for 16t^2=20t equation:

16t^2=20t

We move all terms to the left:

16t^2-(20t)=0

a = 16; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·16·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*16}=\frac{0}{32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*16}=\frac{40}{32}
=1+1/4
$