(2x^2)+19x-4.7=0

Solution for (2x^2)+19x-4.7=0 equation:

(2x^2)+19x-4.7=0

a = 2; b = 19; c = -4.7;
Δ = b2-4ac
Δ = 192-4·2·(-4.7)
Δ = 398.6
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{398.6}}{2*2}=\frac{-19-\sqrt{398.6}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{398.6}}{2*2}=\frac{-19+\sqrt{398.6}}{4}
$