3x+4(x2)=6x-1+8(2x+3)

Solution for 3x+4(x2)=6x-1+8(2x+3) equation:

3x+4(x2)=6x-1+8(2x+3)

We move all terms to the left:

3x+4(x2)-(6x-1+8(2x+3))=0

We add all the numbers together, and all the variables

4x^2+3x-(6x-1+8(2x+3))=0


We calculate terms in parentheses: -(6x-1+8(2x+3)), so:

6x-1+8(2x+3)

determiningTheFunctionDomain
6x+8(2x+3)-1

We multiply parentheses

6x+16x+24-1

We add all the numbers together, and all the variables

22x+23

Back to the equation:

-(22x+23)


We get rid of parentheses

4x^2+3x-22x-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{-8}{8}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+27}{2*4}=\frac{46}{8}
=5+3/4
$