4.9t^2+7t=0

Solution for 4.9t^2+7t=0 equation:

4.9t^2+7t=0

a = 4.9; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·4.9·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*4.9}=\frac{-14}{9.8}
=-1+3.5/8.16666666667
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*4.9}=\frac{0}{9.8}
=0
$