3(x-1)+x2=3(x-1)+2

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

3(x-1)+x2=3(x-1)+2

We move all terms to the left:

3(x-1)+x2-(3(x-1)+2)=0

We add all the numbers together, and all the variables

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

We multiply parentheses

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


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

3(x-1)+2

We multiply parentheses

3x-3+2

We add all the numbers together, and all the variables

3x-1

Back to the equation:

-(3x-1)


We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-2=0

a = 1; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·1·(-2)
Δ = 8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{2}}{2*1}=\frac{0-2\sqrt{2}}{2}
=-\frac{2\sqrt{2}}{2}
=-\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{2}}{2*1}=\frac{0+2\sqrt{2}}{2}
=\frac{2\sqrt{2}}{2}
=\sqrt{2}
$