2x(3+x)3=7x(2+x)2

Solution for 2x(3+x)3=7x(2+x)2 equation:

2x(3+x)3=7x(2+x)2

We move all terms to the left:

2x(3+x)3-(7x(2+x)2)=0

We add all the numbers together, and all the variables

2x(x+3)3-(7x(x+2)2)=0

We multiply parentheses

6x^2+18x-(7x(x+2)2)=0


We calculate terms in parentheses: -(7x(x+2)2), so:

7x(x+2)2

We multiply parentheses

14x^2+28x

Back to the equation:

-(14x^2+28x)


We get rid of parentheses

6x^2-14x^2+18x-28x=0

We add all the numbers together, and all the variables

-8x^2-10x=0

a = -8; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·(-8)·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2*-8}=\frac{0}{-16}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*-8}=\frac{20}{-16}
=-1+1/4
$