2x+47=3x*(x+4)

Solution for 2x+47=3x*(x+4) equation:

2x+47=3x(x+4)

We move all terms to the left:

2x+47-(3x(x+4))=0


We calculate terms in parentheses: -(3x(x+4)), so:

3x(x+4)

We multiply parentheses

3x^2+12x

Back to the equation:

-(3x^2+12x)


We get rid of parentheses

-3x^2+2x-12x+47=0

We add all the numbers together, and all the variables

-3x^2-10x+47=0

a = -3; b = -10; c = +47;
Δ = b2-4ac
Δ = -102-4·(-3)·47
Δ = 664
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:

$\sqrt{\Delta}=\sqrt{664}=\sqrt{4*166}=\sqrt{4}*\sqrt{166}=2\sqrt{166}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{166}}{2*-3}=\frac{10-2\sqrt{166}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{166}}{2*-3}=\frac{10+2\sqrt{166}}{-6}
$