Solve x(x+1)+2(x-1)=7

$$x ( x + 1 ) + 2 ( x - 1 ) = 7$$

Solve for x

$$x = \frac{3 \sqrt{5} - 3}{2} \approx 1.854101966$$

$$x=\frac{-3\sqrt{5}-3}{2}\approx -4.854101966$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x by x+1.

$$x^{2}+x+2\left(x-1\right)=7$$

Use the distributive property to multiply 2 by x-1.

$$x^{2}+x+2x-2=7$$

Combine x and 2x to get 3x.

$$x^{2}+3x-2=7$$

Subtract 7 from both sides.

$$x^{2}+3x-2-7=0$$

Subtract 7 from -2 to get -9.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-9\right)}}{2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-9\right)}}{2}$$

Multiply -4 times -9.

$$x=\frac{-3±\sqrt{9+36}}{2}$$

Add 9 to 36.

$$x=\frac{-3±\sqrt{45}}{2}$$

Take the square root of 45.

$$x=\frac{-3±3\sqrt{5}}{2}$$

Now solve the equation x=\frac{-3±3\sqrt{5}}{2} when ± is plus. Add -3 to 3\sqrt{5}.

$$x=\frac{3\sqrt{5}-3}{2}$$

Now solve the equation x=\frac{-3±3\sqrt{5}}{2} when ± is minus. Subtract 3\sqrt{5} from -3.

$$x=\frac{-3\sqrt{5}-3}{2}$$

The equation is now solved.

$$x=\frac{3\sqrt{5}-3}{2} x=\frac{-3\sqrt{5}-3}{2}$$

Steps for Completing the Square

Use the distributive property to multiply x by x+1.

$$x^{2}+x+2\left(x-1\right)=7$$

Use the distributive property to multiply 2 by x-1.

$$x^{2}+x+2x-2=7$$

Combine x and 2x to get 3x.

$$x^{2}+3x-2=7$$

Add 2 to both sides.

$$x^{2}+3x=7+2$$

Add 7 and 2 to get 9.

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

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

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=9+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=9+\frac{9}{4}$$

Add 9 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{45}{4}$$

Factor x^{2}+3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{45}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{45}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{3\sqrt{5}}{2} x+\frac{3}{2}=-\frac{3\sqrt{5}}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{3\sqrt{5}-3}{2} x=\frac{-3\sqrt{5}-3}{2}$$