(k+7)(k-3)=(k-3)

Solution for (k+7)(k-3)=(k-3) equation:

(k+7)(k-3)=(k-3)

We move all terms to the left:

(k+7)(k-3)-((k-3))=0

We multiply parentheses ..

(+k^2-3k+7k-21)-((k-3))=0


We calculate terms in parentheses: -((k-3)), so:

(k-3)

We get rid of parentheses

k-3

Back to the equation:

-(k-3)


We get rid of parentheses

k^2-3k+7k-k-21+3=0

We add all the numbers together, and all the variables

k^2+3k-18=0

a = 1; b = 3; c = -18;
Δ = b2-4ac
Δ = 32-4·1·(-18)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*1}=\frac{-12}{2}
=-6
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*1}=\frac{6}{2}
=3
$