-2n^2-3n+8=3n^2+4

Solution for -2n^2-3n+8=3n^2+4 equation:

-2n^2-3n+8=3n^2+4

We move all terms to the left:

-2n^2-3n+8-(3n^2+4)=0

We get rid of parentheses

-2n^2-3n^2-3n-4+8=0

We add all the numbers together, and all the variables

-5n^2-3n+4=0

a = -5; b = -3; c = +4;
Δ = b2-4ac
Δ = -32-4·(-5)·4
Δ = 89
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{89}}{2*-5}=\frac{3-\sqrt{89}}{-10}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{89}}{2*-5}=\frac{3+\sqrt{89}}{-10}
$