To find equation solutions, solve $x=0$ and $7x-42=0$.
$$x=0$$ $$x=6$$
Steps Using the Quadratic Formula
Subtract $42x$ from both sides.
$$7x^{2}-42x=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $7$ for $a$, $-42$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Now solve the equation $x=\frac{42±42}{14}$ when $±$ is plus. Add $42$ to $42$.
$$x=\frac{84}{14}$$
Divide $84$ by $14$.
$$x=6$$
Now solve the equation $x=\frac{42±42}{14}$ when $±$ is minus. Subtract $42$ from $42$.
$$x=\frac{0}{14}$$
Divide $0$ by $14$.
$$x=0$$
The equation is now solved.
$$x=6$$ $$x=0$$
Steps for Completing the Square
Subtract $42x$ from both sides.
$$7x^{2}-42x=0$$
Divide both sides by $7$.
$$\frac{7x^{2}-42x}{7}=\frac{0}{7}$$
Dividing by $7$ undoes the multiplication by $7$.
$$x^{2}+\left(-\frac{42}{7}\right)x=\frac{0}{7}$$
Divide $-42$ by $7$.
$$x^{2}-6x=\frac{0}{7}$$
Divide $0$ by $7$.
$$x^{2}-6x=0$$
Divide $-6$, the coefficient of the $x$ term, by $2$ to get $-3$. Then add the square of $-3$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
