Solve 7x^2=42x

$$7 { x }^{ 2 } =42x$$

Solve for x

$$x=6$$

$$x=0$$

Steps Using Factoring

Subtract 42x from both sides.

$$7x^{2}-42x=0$$

Factor out x.

$$x\left(7x-42\right)=0$$

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}.

$$x=\frac{-\left(-42\right)±\sqrt{\left(-42\right)^{2}}}{2\times 7}$$

Take the square root of \left(-42\right)^{2}.

$$x=\frac{-\left(-42\right)±42}{2\times 7}$$

The opposite of -42 is 42.

$$x=\frac{42±42}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{42±42}{14}$$

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.

$$x^{2}-6x+\left(-3\right)^{2}=\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=9$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{9}$$

Simplify.

$$x-3=3 x-3=-3$$

Add 3 to both sides of the equation.

$$x=6 x=0$$