2x(2+3x)=22(4+x)

Solution for 2x(2+3x)=22(4+x) equation:

2x(2+3x)=22(4+x)

We move all terms to the left:

2x(2+3x)-(22(4+x))=0

We add all the numbers together, and all the variables

2x(3x+2)-(22(x+4))=0

We multiply parentheses

6x^2+4x-(22(x+4))=0


We calculate terms in parentheses: -(22(x+4)), so:

22(x+4)

We multiply parentheses

22x+88

Back to the equation:

-(22x+88)


We get rid of parentheses

6x^2+4x-22x-88=0

We add all the numbers together, and all the variables

6x^2-18x-88=0

a = 6; b = -18; c = -88;
Δ = b2-4ac
Δ = -182-4·6·(-88)
Δ = 2436
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:

$\sqrt{\Delta}=\sqrt{2436}=\sqrt{4*609}=\sqrt{4}*\sqrt{609}=2\sqrt{609}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{609}}{2*6}=\frac{18-2\sqrt{609}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{609}}{2*6}=\frac{18+2\sqrt{609}}{12}
$