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4x-4=x^2+(x+1)(x-1)
We move all terms to the left:
4x-4-(x^2+(x+1)(x-1))=0
We use the square of the difference formula
x^2+4x+1-4=0
We add all the numbers together, and all the variables
x^2+4x-3=0
a = 1; b = 4; c = -3;
Δ = b2-4ac
Δ = 42-4·1·(-3)
Δ = 28
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{7}}{2*1}=\frac{-4-2\sqrt{7}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{7}}{2*1}=\frac{-4+2\sqrt{7}}{2} $
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