(x+2)(3x)=(x+2)(21)

Solution for (x+2)(3x)=(x+2)(21) equation:

(x+2)(3x)=(x+2)(21)

We move all terms to the left:

(x+2)(3x)-((x+2)(21))=0

We multiply parentheses

3x^2+6x-((x+2)21)=0


We calculate terms in parentheses: -((x+2)21), so:

(x+2)21

We multiply parentheses

21x+42

Back to the equation:

-(21x+42)


We get rid of parentheses

3x^2+6x-21x-42=0

We add all the numbers together, and all the variables

3x^2-15x-42=0

a = 3; b = -15; c = -42;
Δ = b2-4ac
Δ = -152-4·3·(-42)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-27}{2*3}=\frac{-12}{6}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+27}{2*3}=\frac{42}{6}
=7
$