3x+18=2(x+9)x

Solution for 3x+18=2(x+9)x equation:

3x+18=2(x+9)x

We move all terms to the left:

3x+18-(2(x+9)x)=0


We calculate terms in parentheses: -(2(x+9)x), so:

2(x+9)x

We multiply parentheses

2x^2+18x

Back to the equation:

-(2x^2+18x)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-2x^2-15x+18=0

a = -2; b = -15; c = +18;
Δ = b2-4ac
Δ = -152-4·(-2)·18
Δ = 369
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:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3\sqrt{41}}{2*-2}=\frac{15-3\sqrt{41}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3\sqrt{41}}{2*-2}=\frac{15+3\sqrt{41}}{-4}
$