-49t^2+150t=0

Solution for -49t^2+150t=0 equation:

-49t^2+150t=0

a = -49; b = 150; c = 0;
Δ = b2-4ac
Δ = 1502-4·(-49)·0
Δ = 22500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{22500}=150$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(150)-150}{2*-49}=\frac{-300}{-98}
=3+3/49
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(150)+150}{2*-49}=\frac{0}{-98}
=0
$