2x^2+18x-19=0

Solution for 2x^2+18x-19=0 equation:

2x^2+18x-19=0

a = 2; b = 18; c = -19;
Δ = b2-4ac
Δ = 182-4·2·(-19)
Δ = 476
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:

$\sqrt{\Delta}=\sqrt{476}=\sqrt{4*119}=\sqrt{4}*\sqrt{119}=2\sqrt{119}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{119}}{2*2}=\frac{-18-2\sqrt{119}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{119}}{2*2}=\frac{-18+2\sqrt{119}}{4}
$