Solve (x+4)(2x-3)=6

$$( x + 4 ) ( 2 x - 3 ) = 6$$

Solve for x

$$x = -\frac{9}{2} = -4\frac{1}{2} = -4.5$$

$$x=2$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x+4 by 2x-3 and combine like terms.

$$2x^{2}+5x-12=6$$

Subtract 6 from both sides.

$$2x^{2}+5x-12-6=0$$

Subtract 6 from -12 to get -18.

$$2x^{2}+5x-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 5 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 2\left(-18\right)}}{2\times 2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 2\left(-18\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-5±\sqrt{25-8\left(-18\right)}}{2\times 2}$$

Multiply -8 times -18.

$$x=\frac{-5±\sqrt{25+144}}{2\times 2}$$

Add 25 to 144.

$$x=\frac{-5±\sqrt{169}}{2\times 2}$$

Take the square root of 169.

$$x=\frac{-5±13}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-5±13}{4}$$

Now solve the equation x=\frac{-5±13}{4} when ± is plus. Add -5 to 13.

$$x=\frac{8}{4}$$

Divide 8 by 4.

$$x=2$$

Now solve the equation x=\frac{-5±13}{4} when ± is minus. Subtract 13 from -5.

$$x=-\frac{18}{4}$$

Reduce the fraction \frac{-18}{4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{9}{2}$$

The equation is now solved.

$$x=2 x=-\frac{9}{2}$$

Steps for Completing the Square

Use the distributive property to multiply x+4 by 2x-3 and combine like terms.

$$2x^{2}+5x-12=6$$

Add 12 to both sides.

$$2x^{2}+5x=6+12$$

Add 6 and 12 to get 18.

$$2x^{2}+5x=18$$

Divide both sides by 2.

$$\frac{2x^{2}+5x}{2}=\frac{18}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{5}{2}x=\frac{18}{2}$$

Divide 18 by 2.

$$x^{2}+\frac{5}{2}x=9$$

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.

$$x^{2}+\frac{5}{2}x+\left(\frac{5}{4}\right)^{2}=9+\left(\frac{5}{4}\right)^{2}$$

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

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=9+\frac{25}{16}$$

Add 9 to \frac{25}{16}.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{169}{16}$$

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}.

$$\left(x+\frac{5}{4}\right)^{2}=\frac{169}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{\frac{169}{16}}$$

Simplify.

$$x+\frac{5}{4}=\frac{13}{4} x+\frac{5}{4}=-\frac{13}{4}$$

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

$$x=2 x=-\frac{9}{2}$$