Solve x*(6x-7)=3

$$x \cdot ( 6 x - 7 ) = 3$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$x=-\frac{1}{3}\approx -0.333333333$$

Steps Using the Quadratic Formula

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

$$6x^{2}-7x=3$$

Subtract 3 from both sides.

$$6x^{2}-7x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -7 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 6\left(-3\right)}}{2\times 6}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 6\left(-3\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-7\right)±\sqrt{49-24\left(-3\right)}}{2\times 6}$$

Multiply -24 times -3.

$$x=\frac{-\left(-7\right)±\sqrt{49+72}}{2\times 6}$$

Add 49 to 72.

$$x=\frac{-\left(-7\right)±\sqrt{121}}{2\times 6}$$

Take the square root of 121.

$$x=\frac{-\left(-7\right)±11}{2\times 6}$$

The opposite of -7 is 7.

$$x=\frac{7±11}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{7±11}{12}$$

Now solve the equation x=\frac{7±11}{12} when ± is plus. Add 7 to 11.

$$x=\frac{18}{12}$$

Reduce the fraction \frac{18}{12} to lowest terms by extracting and canceling out 6.

$$x=\frac{3}{2}$$

Now solve the equation x=\frac{7±11}{12} when ± is minus. Subtract 11 from 7.

$$x=-\frac{4}{12}$$

Reduce the fraction \frac{-4}{12} to lowest terms by extracting and canceling out 4.

$$x=-\frac{1}{3}$$

The equation is now solved.

$$x=\frac{3}{2} x=-\frac{1}{3}$$

Steps for Completing the Square

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

$$6x^{2}-7x=3$$

Divide both sides by 6.

$$\frac{6x^{2}-7x}{6}=\frac{3}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}-\frac{7}{6}x=\frac{3}{6}$$

Reduce the fraction \frac{3}{6} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{7}{6}x=\frac{1}{2}$$

Divide -\frac{7}{6}, the coefficient of the x term, by 2 to get -\frac{7}{12}. Then add the square of -\frac{7}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{6}x+\left(-\frac{7}{12}\right)^{2}=\frac{1}{2}+\left(-\frac{7}{12}\right)^{2}$$

Square -\frac{7}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{6}x+\frac{49}{144}=\frac{1}{2}+\frac{49}{144}$$

Add \frac{1}{2} to \frac{49}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{6}x+\frac{49}{144}=\frac{121}{144}$$

Factor x^{2}-\frac{7}{6}x+\frac{49}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{12}\right)^{2}=\frac{121}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{12}\right)^{2}}=\sqrt{\frac{121}{144}}$$

Simplify.

$$x-\frac{7}{12}=\frac{11}{12} x-\frac{7}{12}=-\frac{11}{12}$$

Add \frac{7}{12} to both sides of the equation.

$$x=\frac{3}{2} x=-\frac{1}{3}$$