4x(x+2)+x+1=2x-3+3(x+4)

Solution for 4x(x+2)+x+1=2x-3+3(x+4) equation:

4x(x+2)+x+1=2x-3+3(x+4)

We move all terms to the left:

4x(x+2)+x+1-(2x-3+3(x+4))=0

We add all the numbers together, and all the variables

x+4x(x+2)-(2x-3+3(x+4))+1=0

We multiply parentheses

4x^2+x+8x-(2x-3+3(x+4))+1=0


We calculate terms in parentheses: -(2x-3+3(x+4)), so:

2x-3+3(x+4)

determiningTheFunctionDomain
2x+3(x+4)-3

We multiply parentheses

2x+3x+12-3

We add all the numbers together, and all the variables

5x+9

Back to the equation:

-(5x+9)


We add all the numbers together, and all the variables

4x^2+9x-(5x+9)+1=0

We get rid of parentheses

4x^2+9x-5x-9+1=0

We add all the numbers together, and all the variables

4x^2+4x-8=0

a = 4; b = 4; c = -8;
Δ = b2-4ac
Δ = 42-4·4·(-8)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12}{2*4}=\frac{-16}{8}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*4}=\frac{8}{8}
=1
$