4.9t^2+1.5t-57=0

Solution for 4.9t^2+1.5t-57=0 equation:

4.9t^2+1.5t-57=0

a = 4.9; b = 1.5; c = -57;
Δ = b2-4ac
Δ = 1.52-4·4.9·(-57)
Δ = 1119.45
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{-(1.5)-\sqrt{1119.45}}{2*4.9}=\frac{-1.5-\sqrt{1119.45}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.5)+\sqrt{1119.45}}{2*4.9}=\frac{-1.5+\sqrt{1119.45}}{9.8}
$