(2x+3)(4+x)=9(x+3)+x

Solution for (2x+3)(4+x)=9(x+3)+x equation:

(2x+3)(4+x)=9(x+3)+x

We move all terms to the left:

(2x+3)(4+x)-(9(x+3)+x)=0

We add all the numbers together, and all the variables

(2x+3)(x+4)-(9(x+3)+x)=0

We multiply parentheses ..

(+2x^2+8x+3x+12)-(9(x+3)+x)=0


We calculate terms in parentheses: -(9(x+3)+x), so:

9(x+3)+x

We add all the numbers together, and all the variables

x+9(x+3)

We multiply parentheses

x+9x+27

We add all the numbers together, and all the variables

10x+27

Back to the equation:

-(10x+27)


We get rid of parentheses

2x^2+8x+3x-10x+12-27=0

We add all the numbers together, and all the variables

2x^2+x-15=0

a = 2; b = 1; c = -15;
Δ = b2-4ac
Δ = 12-4·2·(-15)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-11}{2*2}=\frac{-12}{4}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*2}=\frac{10}{4}
=2+1/2
$