4n-3n^2=-6

Solution for 4n-3n^2=-6 equation:

4n-3n^2=-6

We move all terms to the left:

4n-3n^2-(-6)=0

We add all the numbers together, and all the variables

-3n^2+4n+6=0

a = -3; b = 4; c = +6;
Δ = b2-4ac
Δ = 42-4·(-3)·6
Δ = 88
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{22}}{2*-3}=\frac{-4-2\sqrt{22}}{-6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{22}}{2*-3}=\frac{-4+2\sqrt{22}}{-6}
$