-0.7t^2+24t=0

Solution for -0.7t^2+24t=0 equation:

-0.7t^2+24t=0

a = -0.7; b = 24; c = 0;
Δ = b2-4ac
Δ = 242-4·(-0.7)·0
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-24}{2*-0.7}=\frac{-48}{-1.4}
=34+0.4/1.4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+24}{2*-0.7}=\frac{0}{-1.4}
=0
$