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

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

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

We move all terms to the left:

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

We get rid of parentheses

x^2-(2x-1)/3-2x-4=0

We multiply all the terms by the denominator


x^2*3-(2x-1)-2x*3-4*3=0

We add all the numbers together, and all the variables

x^2*3-(2x-1)-2x*3-12=0

Wy multiply elements

3x^2-(2x-1)-6x-12=0

We get rid of parentheses

3x^2-2x-6x+1-12=0

We add all the numbers together, and all the variables

3x^2-8x-11=0

a = 3; b = -8; c = -11;
Δ = b2-4ac
Δ = -82-4·3·(-11)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-14}{2*3}=\frac{-6}{6}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+14}{2*3}=\frac{22}{6}
=3+2/3
$