Solve (8x+1)(x-2)=13x

$$( 8 x + 1 ) ( x - 2 ) = 13 x$$

Solve for x

$$x = \frac{\sqrt{53} + 7}{4} \approx 3.570027472$$

$$x=\frac{7-\sqrt{53}}{4}\approx -0.070027472$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 8x+1 by x-2 and combine like terms.

$$8x^{2}-15x-2=13x$$

Subtract 13x from both sides.

$$8x^{2}-15x-2-13x=0$$

Combine -15x and -13x to get -28x.

$$8x^{2}-28x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -28 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 8\left(-2\right)}}{2\times 8}$$

Square -28.

$$x=\frac{-\left(-28\right)±\sqrt{784-4\times 8\left(-2\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-28\right)±\sqrt{784-32\left(-2\right)}}{2\times 8}$$

Multiply -32 times -2.

$$x=\frac{-\left(-28\right)±\sqrt{784+64}}{2\times 8}$$

Add 784 to 64.

$$x=\frac{-\left(-28\right)±\sqrt{848}}{2\times 8}$$

Take the square root of 848.

$$x=\frac{-\left(-28\right)±4\sqrt{53}}{2\times 8}$$

The opposite of -28 is 28.

$$x=\frac{28±4\sqrt{53}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{28±4\sqrt{53}}{16}$$

Now solve the equation x=\frac{28±4\sqrt{53}}{16} when ± is plus. Add 28 to 4\sqrt{53}.

$$x=\frac{4\sqrt{53}+28}{16}$$

Divide 28+4\sqrt{53} by 16.

$$x=\frac{\sqrt{53}+7}{4}$$

Now solve the equation x=\frac{28±4\sqrt{53}}{16} when ± is minus. Subtract 4\sqrt{53} from 28.

$$x=\frac{28-4\sqrt{53}}{16}$$

Divide 28-4\sqrt{53} by 16.

$$x=\frac{7-\sqrt{53}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{53}+7}{4} x=\frac{7-\sqrt{53}}{4}$$

Steps for Completing the Square

Use the distributive property to multiply 8x+1 by x-2 and combine like terms.

$$8x^{2}-15x-2=13x$$

Subtract 13x from both sides.

$$8x^{2}-15x-2-13x=0$$

Combine -15x and -13x to get -28x.

$$8x^{2}-28x-2=0$$

Add 2 to both sides. Anything plus zero gives itself.

$$8x^{2}-28x=2$$

Divide both sides by 8.

$$\frac{8x^{2}-28x}{8}=\frac{2}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\left(-\frac{28}{8}\right)x=\frac{2}{8}$$

Reduce the fraction \frac{-28}{8} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{7}{2}x=\frac{2}{8}$$

Reduce the fraction \frac{2}{8} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{7}{2}x=\frac{1}{4}$$

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=\frac{1}{4}+\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{1}{4}+\frac{49}{16}$$

Add \frac{1}{4} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{53}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{53}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{53}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{\sqrt{53}}{4} x-\frac{7}{4}=-\frac{\sqrt{53}}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$x=\frac{\sqrt{53}+7}{4} x=\frac{7-\sqrt{53}}{4}$$