(x+2)+(x+4)(x+6)=4(x+18)

Solution for (x+2)+(x+4)(x+6)=4(x+18) equation:

(x+2)+(x+4)(x+6)=4(x+18)

We move all terms to the left:

(x+2)+(x+4)(x+6)-(4(x+18))=0

We get rid of parentheses

x+(x+4)(x+6)-(4(x+18))+2=0

We multiply parentheses ..

(+x^2+6x+4x+24)+x-(4(x+18))+2=0


We calculate terms in parentheses: -(4(x+18)), so:

4(x+18)

We multiply parentheses

4x+72

Back to the equation:

-(4x+72)


We get rid of parentheses

x^2+6x+4x+x-4x+24-72+2=0

We add all the numbers together, and all the variables

x^2+7x-46=0

a = 1; b = 7; c = -46;
Δ = b2-4ac
Δ = 72-4·1·(-46)
Δ = 233
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{233}}{2*1}=\frac{-7-\sqrt{233}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{233}}{2*1}=\frac{-7+\sqrt{233}}{2}
$