x2-x-1=(x+2)(2x+3)

Solution for x2-x-1=(x+2)(2x+3) equation:

x2-x-1=(x+2)(2x+3)

We move all terms to the left:

x2-x-1-((x+2)(2x+3))=0

We add all the numbers together, and all the variables

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

We multiply parentheses ..

x^2-((+2x^2+3x+4x+6))-1x-1=0


We calculate terms in parentheses: -((+2x^2+3x+4x+6)), so:

(+2x^2+3x+4x+6)

We get rid of parentheses

2x^2+3x+4x+6

We add all the numbers together, and all the variables

2x^2+7x+6

Back to the equation:

-(2x^2+7x+6)


We add all the numbers together, and all the variables

x^2-1x-(2x^2+7x+6)-1=0

We get rid of parentheses

x^2-2x^2-1x-7x-6-1=0

We add all the numbers together, and all the variables

-1x^2-8x-7=0

a = -1; b = -8; c = -7;
Δ = b2-4ac
Δ = -82-4·(-1)·(-7)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-6}{2*-1}=\frac{2}{-2}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+6}{2*-1}=\frac{14}{-2}
=-7
$