Answer
The solutions are $x=4$ and $x=3$.
Work Step by Step
The given equation is
$\Rightarrow x^2-7x+12=0$
The equation is easily factorable. So, solve by factoring.
$\Rightarrow x^2-7x+12=0$
Rewrite $-7x$ as $-4x-3x$.
$\Rightarrow x^2-4x-3x+12=0$
Group the terms.
$\Rightarrow (x^2-4x)+(-3x+12)=0$
Factor each group.
$\Rightarrow x(x-4)-3(x-4)=0$
Factor out $(x-4)$.
$\Rightarrow (x-4)(x-3)=0$
Use zero-product property.
$\Rightarrow x-4=0$ or $x-3=0$
Solve for $x$.
$\Rightarrow x=4$ or $x=3$
Hence, the solutions are $x=4$ and $x=3$.
