2x(2x+3)-7x=23-5(x+3x)

Solution for 2x(2x+3)-7x=23-5(x+3x) equation:

2x(2x+3)-7x=23-5(x+3x)

We move all terms to the left:

2x(2x+3)-7x-(23-5(x+3x))=0

We add all the numbers together, and all the variables

2x(2x+3)-7x-(23-5(+4x))=0

We add all the numbers together, and all the variables

-7x+2x(2x+3)-(23-5(+4x))=0

We multiply parentheses

4x^2-7x+6x-(23-5(+4x))=0


We calculate terms in parentheses: -(23-5(+4x)), so:

23-5(+4x)

determiningTheFunctionDomain
-5(+4x)+23

We multiply parentheses

-20x+23

Back to the equation:

-(-20x+23)


We add all the numbers together, and all the variables

4x^2-1x-(-20x+23)=0

We get rid of parentheses

4x^2-1x+20x-23=0

We add all the numbers together, and all the variables

4x^2+19x-23=0

a = 4; b = 19; c = -23;
Δ = b2-4ac
Δ = 192-4·4·(-23)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-27}{2*4}=\frac{-46}{8}
=-5+3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+27}{2*4}=\frac{8}{8}
=1
$