4.9x^2+19x-50=0

Solution for 4.9x^2+19x-50=0 equation:

4.9x^2+19x-50=0

a = 4.9; b = 19; c = -50;
Δ = b2-4ac
Δ = 192-4·4.9·(-50)
Δ = 1341
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1341}=\sqrt{9*149}=\sqrt{9}*\sqrt{149}=3\sqrt{149}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-3\sqrt{149}}{2*4.9}=\frac{-19-3\sqrt{149}}{9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+3\sqrt{149}}{2*4.9}=\frac{-19+3\sqrt{149}}{9.8}
$