26t-3.2t^2=0

Solution for 26t-3.2t^2=0 equation:

26t-3.2t^2=0

a = -3.2; b = 26; c = 0;
Δ = b2-4ac
Δ = 262-4·(-3.2)·0
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-26}{2*-3.2}=\frac{-52}{-6.4}
=8+0.8/6.4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+26}{2*-3.2}=\frac{0}{-6.4}
=0
$