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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+8x^2-28x+6x-21)-4=0

We get rid of parentheses

8x^2-28x+6x-21-4=0

We add all the numbers together, and all the variables

8x^2-22x-25=0

a = 8; b = -22; c = -25;
Δ = b2-4ac
Δ = -222-4·8·(-25)
Δ = 1284
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:

$\sqrt{\Delta}=\sqrt{1284}=\sqrt{4*321}=\sqrt{4}*\sqrt{321}=2\sqrt{321}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{321}}{2*8}=\frac{22-2\sqrt{321}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{321}}{2*8}=\frac{22+2\sqrt{321}}{16}
$