5.7t-4.9t^2=0

Solution for 5.7t-4.9t^2=0 equation:

5.7t-4.9t^2=0

a = -4.9; b = 5.7; c = 0;
Δ = b2-4ac
Δ = 5.72-4·(-4.9)·0
Δ = 32.49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5.7)-\sqrt{32.49}}{2*-4.9}=\frac{-5.7-\sqrt{32.49}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5.7)+\sqrt{32.49}}{2*-4.9}=\frac{-5.7+\sqrt{32.49}}{-9.8}
$