Solve (3x-4)(x+1)=20

$$( 3 x - 4 ) ( x + 1 ) = 20$$

Solve for x

$$x = -\frac{8}{3} = -2\frac{2}{3} \approx -2.666666667$$

$$x=3$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x-4 by x+1 and combine like terms.

$$3x^{2}-x-4=20$$

Subtract 20 from both sides.

$$3x^{2}-x-4-20=0$$

Subtract 20 from -4 to get -24.

$$3x^{2}-x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -1 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 3\left(-24\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-1\right)±\sqrt{1-12\left(-24\right)}}{2\times 3}$$

Multiply -12 times -24.

$$x=\frac{-\left(-1\right)±\sqrt{1+288}}{2\times 3}$$

Add 1 to 288.

$$x=\frac{-\left(-1\right)±\sqrt{289}}{2\times 3}$$

Take the square root of 289.

$$x=\frac{-\left(-1\right)±17}{2\times 3}$$

The opposite of -1 is 1.

$$x=\frac{1±17}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{1±17}{6}$$

Now solve the equation x=\frac{1±17}{6} when ± is plus. Add 1 to 17.

$$x=\frac{18}{6}$$

Divide 18 by 6.

$$x=3$$

Now solve the equation x=\frac{1±17}{6} when ± is minus. Subtract 17 from 1.

$$x=-\frac{16}{6}$$

Reduce the fraction \frac{-16}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{8}{3}$$

The equation is now solved.

$$x=3 x=-\frac{8}{3}$$

Steps for Completing the Square

Use the distributive property to multiply 3x-4 by x+1 and combine like terms.

$$3x^{2}-x-4=20$$

Add 4 to both sides.

$$3x^{2}-x=20+4$$

Add 20 and 4 to get 24.

$$3x^{2}-x=24$$

Divide both sides by 3.

$$\frac{3x^{2}-x}{3}=\frac{24}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{1}{3}x=\frac{24}{3}$$

Divide 24 by 3.

$$x^{2}-\frac{1}{3}x=8$$

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=8+\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=8+\frac{1}{36}$$

Add 8 to \frac{1}{36}.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{289}{36}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{289}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{289}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{17}{6} x-\frac{1}{6}=-\frac{17}{6}$$

Add \frac{1}{6} to both sides of the equation.

$$x=3 x=-\frac{8}{3}$$