(2x+12)=(3x+23)x

Solution for (2x+12)=(3x+23)x equation:

(2x+12)=(3x+23)x

We move all terms to the left:

(2x+12)-((3x+23)x)=0

We get rid of parentheses

2x-((3x+23)x)+12=0


We calculate terms in parentheses: -((3x+23)x), so:

(3x+23)x

We multiply parentheses

3x^2+23x

Back to the equation:

-(3x^2+23x)


We get rid of parentheses

-3x^2+2x-23x+12=0

We add all the numbers together, and all the variables

-3x^2-21x+12=0

a = -3; b = -21; c = +12;
Δ = b2-4ac
Δ = -212-4·(-3)·12
Δ = 585
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:

$\sqrt{\Delta}=\sqrt{585}=\sqrt{9*65}=\sqrt{9}*\sqrt{65}=3\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-3\sqrt{65}}{2*-3}=\frac{21-3\sqrt{65}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+3\sqrt{65}}{2*-3}=\frac{21+3\sqrt{65}}{-6}
$