How to solve quadratic equations by completing the square when the coefficient of x^2 is not 1?

When $a \neq 1$, first factor out $a$ from the quadratic and linear terms, then complete the square inside the parentheses. For $3x^2 + 12x + 5 = 0$: Step 1: Factor out 3 from the $x$-terms $$3(x^2 + 4x) + 5 = 0$$ Step 2: Complete the square inside: add and subtract $(4/2)^2 = 4$ $$3(x^2 + 4x + 4 - 4) + 5 = 0$$ $$3[(x + 2)^2 - 4] + 5 = 0$$ Step 3: Distribute and simplify $$3(x + 2)^2 - 12 + 5 = 0$$ $$3(x + 2)^2 = 7$$ $$(x + 2)^2 = \frac{7}{3}$$ $$x = -2 \pm \sqrt{\frac{7}{3}}$$ For your second example $5x^2 - 20x + 15 = 0$, factor 5: $5(x^2 - 4x) + 15 = 0$, then $5(x - 2)^2 - 20 + 15 = 0$, giving $5(x - 2)^2 = 5$, so $x = 1$ or $x = 3$. The quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ always works and is often faster when $a \neq 1$, but completing the square gives you the same result and builds intuition for deriving the formula itself.