2n(n-4)=23

Solution for 2n(n-4)=23 equation:

2n(n-4)=23

We move all terms to the left:

2n(n-4)-(23)=0

We multiply parentheses

2n^2-8n-23=0

a = 2; b = -8; c = -23;
Δ = b2-4ac
Δ = -82-4·2·(-23)
Δ = 248
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:

$\sqrt{\Delta}=\sqrt{248}=\sqrt{4*62}=\sqrt{4}*\sqrt{62}=2\sqrt{62}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{62}}{2*2}=\frac{8-2\sqrt{62}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{62}}{2*2}=\frac{8+2\sqrt{62}}{4}
$