4(x+1)(x-1)=4(x+3)

Solution for 4(x+1)(x-1)=4(x+3) equation:

4(x+1)(x-1)=4(x+3)

We move all terms to the left:

4(x+1)(x-1)-(4(x+3))=0

We use the square of the difference formula

x^2-(4(x+3))-1=0


We calculate terms in parentheses: -(4(x+3)), so:

4(x+3)

We multiply parentheses

4x+12

Back to the equation:

-(4x+12)


We get rid of parentheses

x^2-4x-12-1=0

We add all the numbers together, and all the variables

x^2-4x-13=0

a = 1; b = -4; c = -13;
Δ = b2-4ac
Δ = -42-4·1·(-13)
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{17}}{2*1}=\frac{4-2\sqrt{17}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{17}}{2*1}=\frac{4+2\sqrt{17}}{2}
$