(2x+3)2+5x(x+2)=(x-1)(x+1)

Solution for (2x+3)2+5x(x+2)=(x-1)(x+1) equation:

(2x+3)2+5x(x+2)=(x-1)(x+1)

We move all terms to the left:

(2x+3)2+5x(x+2)-((x-1)(x+1))=0

We use the square of the difference formula

x^2+(2x+3)2+5x(x+2)+1=0

We multiply parentheses

x^2+5x^2+4x+10x+6+1=0

We add all the numbers together, and all the variables

6x^2+14x+7=0

a = 6; b = 14; c = +7;
Δ = b2-4ac
Δ = 142-4·6·7
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{-(14)-2\sqrt{7}}{2*6}=\frac{-14-2\sqrt{7}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{7}}{2*6}=\frac{-14+2\sqrt{7}}{12}
$