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

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

Solve for x

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

$$x=2$$

Steps Using the Quadratic Formula

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

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

Subtract 3 from both sides.

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

Subtract 3 from -3 to get -6.

$$2x^{2}-x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -1 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 2\left(-6\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-1\right)±\sqrt{1-8\left(-6\right)}}{2\times 2}$$

Multiply -8 times -6.

$$x=\frac{-\left(-1\right)±\sqrt{1+48}}{2\times 2}$$

Add 1 to 48.

$$x=\frac{-\left(-1\right)±\sqrt{49}}{2\times 2}$$

Take the square root of 49.

$$x=\frac{-\left(-1\right)±7}{2\times 2}$$

The opposite of -1 is 1.

$$x=\frac{1±7}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{1±7}{4}$$

Now solve the equation x=\frac{1±7}{4} when ± is plus. Add 1 to 7.

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

Divide 8 by 4.

$$x=2$$

Now solve the equation x=\frac{1±7}{4} when ± is minus. Subtract 7 from 1.

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

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

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

The equation is now solved.

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

Steps for Completing the Square

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

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

Add 3 to both sides.

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

Add 3 and 3 to get 6.

$$2x^{2}-x=6$$

Divide both sides by 2.

$$\frac{2x^{2}-x}{2}=\frac{6}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{1}{2}x=\frac{6}{2}$$

Divide 6 by 2.

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

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.

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

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

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

Add 3 to \frac{1}{16}.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{49}{16}$$

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}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{7}{4} x-\frac{1}{4}=-\frac{7}{4}$$

Add \frac{1}{4} to both sides of the equation.

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