2x(^2)+19x-5.1=0

Solution for 2x(^2)+19x-5.1=0 equation:

2x(^2)+19x-5.1=0

a = 2; b = 19; c = -5.1;
Δ = b2-4ac
Δ = 192-4·2·(-5.1)
Δ = 401.8
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{401.8}}{2*2}=\frac{-19-\sqrt{401.8}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{401.8}}{2*2}=\frac{-19+\sqrt{401.8}}{4}
$