(4x-7)(2x-3)=2x+

Solution for (4x-7)(2x-3)=2x+ equation:

(4x-7)(2x-3)=2x+

We move all terms to the left:

(4x-7)(2x-3)-(2x+)=0

We add all the numbers together, and all the variables

(4x-7)(2x-3)-(+2x)=0

We get rid of parentheses

(4x-7)(2x-3)-2x=0

We multiply parentheses ..

(+8x^2-12x-14x+21)-2x=0

We get rid of parentheses

8x^2-12x-14x-2x+21=0

We add all the numbers together, and all the variables

8x^2-28x+21=0

a = 8; b = -28; c = +21;
Δ = b2-4ac
Δ = -282-4·8·21
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{7}}{2*8}=\frac{28-4\sqrt{7}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{7}}{2*8}=\frac{28+4\sqrt{7}}{16}
$