(x+7)(x-3)=(x+1)2

Solution for (x+7)(x-3)=(x+1)2 equation:

(x+7)(x-3)=(x+1)2

We move all terms to the left:

(x+7)(x-3)-((x+1)2)=0

We multiply parentheses ..

(+x^2-3x+7x-21)-((x+1)2)=0


We calculate terms in parentheses: -((x+1)2), so:

(x+1)2

We multiply parentheses

2x+2

Back to the equation:

-(2x+2)


We get rid of parentheses

x^2-3x+7x-2x-21-2=0

We add all the numbers together, and all the variables

x^2+2x-23=0

a = 1; b = 2; c = -23;
Δ = b2-4ac
Δ = 22-4·1·(-23)
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{6}}{2*1}=\frac{-2-4\sqrt{6}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{6}}{2*1}=\frac{-2+4\sqrt{6}}{2}
$