-49t^2+147t=0

Solution for -49t^2+147t=0 equation:

-49t^2+147t=0

a = -49; b = 147; c = 0;
Δ = b2-4ac
Δ = 1472-4·(-49)·0
Δ = 21609
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{21609}=147$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(147)-147}{2*-49}=\frac{-294}{-98}
=+3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(147)+147}{2*-49}=\frac{0}{-98}
=0
$