N(x)=-0.4x^2+10.4x+11

Solution for N(x)=-0.4x^2+10.4x+11 equation:

(N)=-0.4N^2+10.4N+11

We move all terms to the left:

(N)-(-0.4N^2+10.4N+11)=0

We get rid of parentheses

0.4N^2-10.4N+N-11=0

We add all the numbers together, and all the variables

0.4N^2-9.4N-11=0

a = 0.4; b = -9.4; c = -11;
Δ = b2-4ac
Δ = -9.42-4·0.4·(-11)
Δ = 105.96
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{-(-9.4)-\sqrt{105.96}}{2*0.4}=\frac{9.4-\sqrt{105.96}}{0.8}
$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9.4)+\sqrt{105.96}}{2*0.4}=\frac{9.4+\sqrt{105.96}}{0.8}
$