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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

(x+2)(x+1)

We multiply parentheses ..

(+x^2+x+2x+2)

We get rid of parentheses

x^2+x+2x+2

We add all the numbers together, and all the variables

x^2+3x+2

Back to the equation:

-(x^2+3x+2)


We get rid of parentheses

2x^2-x^2-2x+5x-3x-5-2=0

We add all the numbers together, and all the variables

x^2-7=0

a = 1; b = 0; c = -7;
Δ = b2-4ac
Δ = 02-4·1·(-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{-(0)-2\sqrt{7}}{2*1}=\frac{0-2\sqrt{7}}{2}
=-\frac{2\sqrt{7}}{2}
=-\sqrt{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{7}}{2*1}=\frac{0+2\sqrt{7}}{2}
=\frac{2\sqrt{7}}{2}
=\sqrt{7}
$