-4.9x^2+19x+40=0

Solution for -4.9x^2+19x+40=0 equation:

-4.9x^2+19x+40=0

a = -4.9; b = 19; c = +40;
Δ = b2-4ac
Δ = 192-4·(-4.9)·40
Δ = 1145
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{1145}}{2*-4.9}=\frac{-19-\sqrt{1145}}{-9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{1145}}{2*-4.9}=\frac{-19+\sqrt{1145}}{-9.8}
$