Solve 2x^2+7x(x-6)=0

$$2 x ^ { 2 } + 7 x ( x - 6 ) = 0$$

Solve for x

$$x = \frac{14}{3} = 4\frac{2}{3} \approx 4.666666667$$

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply 7x by x-6.

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

Combine 2x^{2} and 7x^{2} to get 9x^{2}.

$$9x^{2}-42x=0$$

Factor out x.

$$x\left(9x-42\right)=0$$

To find equation solutions, solve x=0 and 9x-42=0.

$$x=0 x=\frac{14}{3}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 7x by x-6.

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

Combine 2x^{2} and 7x^{2} to get 9x^{2}.

$$9x^{2}-42x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -42 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-42\right)±\sqrt{\left(-42\right)^{2}}}{2\times 9}$$

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

$$x=\frac{-\left(-42\right)±42}{2\times 9}$$

The opposite of -42 is 42.

$$x=\frac{42±42}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{42±42}{18}$$

Now solve the equation x=\frac{42±42}{18} when ± is plus. Add 42 to 42.

$$x=\frac{84}{18}$$

Reduce the fraction \frac{84}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{14}{3}$$

Now solve the equation x=\frac{42±42}{18} when ± is minus. Subtract 42 from 42.

$$x=\frac{0}{18}$$

Divide 0 by 18.

$$x=0$$

The equation is now solved.

$$x=\frac{14}{3} x=0$$

Steps for Completing the Square

Use the distributive property to multiply 7x by x-6.

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

Combine 2x^{2} and 7x^{2} to get 9x^{2}.

$$9x^{2}-42x=0$$

Divide both sides by 9.

$$\frac{9x^{2}-42x}{9}=\frac{0}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{42}{9}\right)x=\frac{0}{9}$$

Reduce the fraction \frac{-42}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{14}{3}x=\frac{0}{9}$$

Divide 0 by 9.

$$x^{2}-\frac{14}{3}x=0$$

Divide -\frac{14}{3}, the coefficient of the x term, by 2 to get -\frac{7}{3}. Then add the square of -\frac{7}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{14}{3}x+\left(-\frac{7}{3}\right)^{2}=\left(-\frac{7}{3}\right)^{2}$$

Square -\frac{7}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{14}{3}x+\frac{49}{9}=\frac{49}{9}$$

Factor x^{2}-\frac{14}{3}x+\frac{49}{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}.

$$\left(x-\frac{7}{3}\right)^{2}=\frac{49}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{3}\right)^{2}}=\sqrt{\frac{49}{9}}$$

Simplify.

$$x-\frac{7}{3}=\frac{7}{3} x-\frac{7}{3}=-\frac{7}{3}$$

Add \frac{7}{3} to both sides of the equation.

$$x=\frac{14}{3} x=0$$