Solve (x+3)(2x-1)=9

$$( x + 3 ) ( 2 x - 1 ) = 9$$

Solve for x

$$x=-4$$

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

Steps Using the Quadratic Formula

Use the distributive property to multiply x+3 by 2x-1 and combine like terms.

$$2x^{2}+5x-3=9$$

Subtract 9 from both sides.

$$2x^{2}+5x-3-9=0$$

Subtract 9 from -3 to get -12.

$$2x^{2}+5x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 5 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 2\left(-12\right)}}{2\times 2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 2\left(-12\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-5±\sqrt{25-8\left(-12\right)}}{2\times 2}$$

Multiply -8 times -12.

$$x=\frac{-5±\sqrt{25+96}}{2\times 2}$$

Add 25 to 96.

$$x=\frac{-5±\sqrt{121}}{2\times 2}$$

Take the square root of 121.

$$x=\frac{-5±11}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-5±11}{4}$$

Now solve the equation x=\frac{-5±11}{4} when ± is plus. Add -5 to 11.

$$x=\frac{6}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-5±11}{4} when ± is minus. Subtract 11 from -5.

$$x=-\frac{16}{4}$$

Divide -16 by 4.

$$x=-4$$

The equation is now solved.

$$x=\frac{3}{2} x=-4$$

Steps for Completing the Square

Use the distributive property to multiply x+3 by 2x-1 and combine like terms.

$$2x^{2}+5x-3=9$$

Add 3 to both sides.

$$2x^{2}+5x=9+3$$

Add 9 and 3 to get 12.

$$2x^{2}+5x=12$$

Divide both sides by 2.

$$\frac{2x^{2}+5x}{2}=\frac{12}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{5}{2}x=\frac{12}{2}$$

Divide 12 by 2.

$$x^{2}+\frac{5}{2}x=6$$

Divide \frac{5}{2}, the coefficient of the x term, by 2 to get \frac{5}{4}. Then add the square of \frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{2}x+\left(\frac{5}{4}\right)^{2}=6+\left(\frac{5}{4}\right)^{2}$$

Square \frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=6+\frac{25}{16}$$

Add 6 to \frac{25}{16}.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{121}{16}$$

Factor x^{2}+\frac{5}{2}x+\frac{25}{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}.

$$\left(x+\frac{5}{4}\right)^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{\frac{121}{16}}$$

Simplify.

$$x+\frac{5}{4}=\frac{11}{4} x+\frac{5}{4}=-\frac{11}{4}$$

Subtract \frac{5}{4} from both sides of the equation.

$$x=\frac{3}{2} x=-4$$