10x+140=(10+x)(x+2)

Solution for 10x+140=(10+x)(x+2) equation:

10x+140=(10+x)(x+2)

We move all terms to the left:

10x+140-((10+x)(x+2))=0

We add all the numbers together, and all the variables

10x-((x+10)(x+2))+140=0

We multiply parentheses ..

-((+x^2+2x+10x+20))+10x+140=0


We calculate terms in parentheses: -((+x^2+2x+10x+20)), so:

(+x^2+2x+10x+20)

We get rid of parentheses

x^2+2x+10x+20

We add all the numbers together, and all the variables

x^2+12x+20

Back to the equation:

-(x^2+12x+20)


We add all the numbers together, and all the variables

10x-(x^2+12x+20)+140=0

We get rid of parentheses

-x^2+10x-12x-20+140=0

We add all the numbers together, and all the variables

-1x^2-2x+120=0

a = -1; b = -2; c = +120;
Δ = b2-4ac
Δ = -22-4·(-1)·120
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-22}{2*-1}=\frac{-20}{-2}
=+10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+22}{2*-1}=\frac{24}{-2}
=-12
$