3x(4+x)=x+18

Solution for 3x(4+x)=x+18 equation:

3x(4+x)=x+18

We move all terms to the left:

3x(4+x)-(x+18)=0

We add all the numbers together, and all the variables

3x(x+4)-(x+18)=0

We multiply parentheses

3x^2+12x-(x+18)=0

We get rid of parentheses

3x^2+12x-x-18=0

We add all the numbers together, and all the variables

3x^2+11x-18=0

a = 3; b = 11; c = -18;
Δ = b2-4ac
Δ = 112-4·3·(-18)
Δ = 337
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{-(11)-\sqrt{337}}{2*3}=\frac{-11-\sqrt{337}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{337}}{2*3}=\frac{-11+\sqrt{337}}{6}
$