x+x(2x+5)=19

Solution for x+x(2x+5)=19 equation:

x+x(2x+5)=19

We move all terms to the left:

x+x(2x+5)-(19)=0

We multiply parentheses

2x^2+x+5x-19=0

We add all the numbers together, and all the variables

2x^2+6x-19=0

a = 2; b = 6; c = -19;
Δ = b2-4ac
Δ = 62-4·2·(-19)
Δ = 188
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:

$\sqrt{\Delta}=\sqrt{188}=\sqrt{4*47}=\sqrt{4}*\sqrt{47}=2\sqrt{47}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{47}}{2*2}=\frac{-6-2\sqrt{47}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{47}}{2*2}=\frac{-6+2\sqrt{47}}{4}
$