N(x)=-0.5x^2+10x+12,

Solution for N(x)=-0.5x^2+10x+12, equation:

(N)=-0.5N^2+10N+12.

We move all terms to the left:

(N)-(-0.5N^2+10N+12.)=0

We get rid of parentheses

0.5N^2-10N+N-12.=0

We add all the numbers together, and all the variables

0.5N^2-9N-12=0

a = 0.5; b = -9; c = -12;
Δ = b2-4ac
Δ = -92-4·0.5·(-12)
Δ = 105
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{-(-9)-\sqrt{105}}{2*0.5}=\frac{9-\sqrt{105}}{1}
$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{105}}{2*0.5}=\frac{9+\sqrt{105}}{1}
$