4.9t^2+5.7t-0.4=0

Solution for 4.9t^2+5.7t-0.4=0 equation:

4.9t^2+5.7t-0.4=0

a = 4.9; b = 5.7; c = -0.4;
Δ = b2-4ac
Δ = 5.72-4·4.9·(-0.4)
Δ = 40.33
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{40.33}}{2*4.9}=\frac{-5.7-\sqrt{40.33}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5.7)+\sqrt{40.33}}{2*4.9}=\frac{-5.7+\sqrt{40.33}}{9.8}
$