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

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

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=-1$$

Steps Using the Quadratic Formula

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

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

Add 3 to both sides.

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

Add -2 and 3 to get 1.

$$2x^{2}+3x+1=0$$

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

$$x=\frac{-3±\sqrt{3^{2}-4\times 2}}{2\times 2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-3±\sqrt{9-8}}{2\times 2}$$

Add 9 to -8.

$$x=\frac{-3±\sqrt{1}}{2\times 2}$$

Take the square root of 1.

$$x=\frac{-3±1}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-3±1}{4}$$

Now solve the equation x=\frac{-3±1}{4} when ± is plus. Add -3 to 1.

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

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

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

Now solve the equation x=\frac{-3±1}{4} when ± is minus. Subtract 1 from -3.

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

Divide -4 by 4.

$$x=-1$$

The equation is now solved.

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

Steps for Completing the Square

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

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

Add 2 to both sides.

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

Add -3 and 2 to get -1.

$$2x^{2}+3x=-1$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=-\frac{1}{2}+\left(\frac{3}{4}\right)^{2}$$

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=-\frac{1}{2}+\frac{9}{16}$$

Add -\frac{1}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{1}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{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}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{1}{4} x+\frac{3}{4}=-\frac{1}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

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