2x^2+4x=19

Solution for 2x^2+4x=19 equation:

2x^2+4x=19

We move all terms to the left:

2x^2+4x-(19)=0

a = 2; b = 4; c = -19;
Δ = b2-4ac
Δ = 42-4·2·(-19)
Δ = 168
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:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{42}}{2*2}=\frac{-4-2\sqrt{42}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{42}}{2*2}=\frac{-4+2\sqrt{42}}{4}
$