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

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

Solve for x

$$x=-1$$

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

Steps Using the Quadratic Formula

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

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

Add 2 to both sides.

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

Add -3 and 2 to get -1.

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

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

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

Square 1.

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

Multiply -4 times 2.

$$x=\frac{-1±\sqrt{1-8\left(-1\right)}}{2\times 2}$$

Multiply -8 times -1.

$$x=\frac{-1±\sqrt{1+8}}{2\times 2}$$

Add 1 to 8.

$$x=\frac{-1±\sqrt{9}}{2\times 2}$$

Take the square root of 9.

$$x=\frac{-1±3}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-1±3}{4}$$

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

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

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{-1±3}{4} when ± is minus. Subtract 3 from -1.

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

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 x-1 by 2x+3 and combine like terms.

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

Add 3 to both sides.

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

Add -2 and 3 to get 1.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

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

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

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

Factor x^{2}+\frac{1}{2}x+\frac{1}{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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

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

Subtract \frac{1}{4} from both sides of the equation.

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