## Precalculus: Mathematics for Calculus, 7th Edition

The real solutions to the equation: $3x^{2}-27=0$ are $x=-3$ and $x=3$
1. $3x^{2}-27=0$ 2. Factor a 3 out of the equation: $3(x^{2}-9)=0$ 3. Divide both sides by 3: $x^{2}-9=0$ 4. Factor the equation using the difference of two squares formula: $a^{2}-b^{2}=(a+b)(a-b)$: $(x+3)(x-3)=0$ 5. Set both factors equal to zero: $x+3=0$ and $x-3=0$ 6. Solve: $x=-3$ and $x=3$