Prove that there are two solutions of the equation $x^2 + kx = 3 - k $
The obvious solution would be finding the value of $b^2 - 4ac$
$x^2 + kx + k - 3=0$
$b^2 - 4 ac$
=$k^2 - 4(1)(k-3)$
=$k^2 - 4k + 12$
From here, $b^2 - 4ac$ have a algebraic value. If it is a constant, I would already be able to prove that the equation have $2$ real roots for every value of "$k$". So, how do I go about doing this question when I can't get $b^2 - 4ac$ to be a constant value?