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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4x^2-1x-3=0

a = 4; b = -1; c = -3;
Δ = b2-4ac
Δ = -12-4·4·(-3)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-7}{2*4}=\frac{-6}{8}
=-3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+7}{2*4}=\frac{8}{8}
=1
$