2x(4x-11)+9=19

Solution for 2x(4x-11)+9=19 equation:

2x(4x-11)+9=19

We move all terms to the left:

2x(4x-11)+9-(19)=0

We add all the numbers together, and all the variables

2x(4x-11)-10=0

We multiply parentheses

8x^2-22x-10=0

a = 8; b = -22; c = -10;
Δ = b2-4ac
Δ = -222-4·8·(-10)
Δ = 804
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:

$\sqrt{\Delta}=\sqrt{804}=\sqrt{4*201}=\sqrt{4}*\sqrt{201}=2\sqrt{201}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{201}}{2*8}=\frac{22-2\sqrt{201}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{201}}{2*8}=\frac{22+2\sqrt{201}}{16}
$