(k+3)(k-3)=40

Solution for (k+3)(k-3)=40 equation:

(k+3)(k-3)=40

We move all terms to the left:

(k+3)(k-3)-(40)=0

We use the square of the difference formula

k^2-9-40=0

We add all the numbers together, and all the variables

k^2-49=0

a = 1; b = 0; c = -49;
Δ = b2-4ac
Δ = 02-4·1·(-49)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-14}{2*1}=\frac{-14}{2}
=-7
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+14}{2*1}=\frac{14}{2}
=7
$