n^2-3n=4-2n^2

Solution for n^2-3n=4-2n^2 equation:

n^2-3n=4-2n^2

We move all terms to the left:

n^2-3n-(4-2n^2)=0

We get rid of parentheses

n^2+2n^2-3n-4=0

We add all the numbers together, and all the variables

3n^2-3n-4=0

a = 3; b = -3; c = -4;
Δ = b2-4ac
Δ = -32-4·3·(-4)
Δ = 57
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{57}}{2*3}=\frac{3-\sqrt{57}}{6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{57}}{2*3}=\frac{3+\sqrt{57}}{6}
$