# How do you find all the zeros of x^2 + 24 = 11x?

$\implies {x}^{2} - 11 x + 24 = 0$
$\implies {x}^{2} - 8 x - 3 x + 24 = 0$
$\implies x \left(x - 8\right) - 3 \left(x - 8\right) = 0$
$\implies \left(x - 8\right) \left(x - 3\right) = 0$
$: . x = 3 \mathmr{and} x = 8$ are the two values for zero