32t-18t^2=0

Solution for 32t-18t^2=0 equation:

32t-18t^2=0

a = -18; b = 32; c = 0;
Δ = b2-4ac
Δ = 322-4·(-18)·0
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32}{2*-18}=\frac{-64}{-36}
=1+7/9
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32}{2*-18}=\frac{0}{-36}
=0
$