(2x-1)(x+4)=(5+x)(2+x)

Solution for (2x-1)(x+4)=(5+x)(2+x) equation:

(2x-1)(x+4)=(5+x)(2+x)

We move all terms to the left:

(2x-1)(x+4)-((5+x)(2+x))=0

We add all the numbers together, and all the variables

(2x-1)(x+4)-((x+5)(x+2))=0

We multiply parentheses ..

(+2x^2+8x-1x-4)-((x+5)(x+2))=0


We calculate terms in parentheses: -((x+5)(x+2)), so:

(x+5)(x+2)

We multiply parentheses ..

(+x^2+2x+5x+10)

We get rid of parentheses

x^2+2x+5x+10

We add all the numbers together, and all the variables

x^2+7x+10

Back to the equation:

-(x^2+7x+10)


We get rid of parentheses

2x^2-x^2+8x-1x-7x-4-10=0

We add all the numbers together, and all the variables

x^2-14=0

a = 1; b = 0; c = -14;
Δ = b2-4ac
Δ = 02-4·1·(-14)
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{14}}{2*1}=\frac{0-2\sqrt{14}}{2}
=-\frac{2\sqrt{14}}{2}
=-\sqrt{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{14}}{2*1}=\frac{0+2\sqrt{14}}{2}
=\frac{2\sqrt{14}}{2}
=\sqrt{14}
$