n(n-7)=28

Solution for n(n-7)=28 equation:

n(n-7)=28

We move all terms to the left:

n(n-7)-(28)=0

We multiply parentheses

n^2-7n-28=0

a = 1; b = -7; c = -28;
Δ = b2-4ac
Δ = -72-4·1·(-28)
Δ = 161
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{-(-7)-\sqrt{161}}{2*1}=\frac{7-\sqrt{161}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{161}}{2*1}=\frac{7+\sqrt{161}}{2}
$