(2x+3)(4x-2)=19

Solution for (2x+3)(4x-2)=19 equation:

(2x+3)(4x-2)=19

We move all terms to the left:

(2x+3)(4x-2)-(19)=0

We multiply parentheses ..

(+8x^2-4x+12x-6)-19=0

We get rid of parentheses

8x^2-4x+12x-6-19=0

We add all the numbers together, and all the variables

8x^2+8x-25=0

a = 8; b = 8; c = -25;
Δ = b2-4ac
Δ = 82-4·8·(-25)
Δ = 864
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:

$\sqrt{\Delta}=\sqrt{864}=\sqrt{144*6}=\sqrt{144}*\sqrt{6}=12\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-12\sqrt{6}}{2*8}=\frac{-8-12\sqrt{6}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+12\sqrt{6}}{2*8}=\frac{-8+12\sqrt{6}}{16}
$