Solve x(x-1)-x(2x+1)=2(1+x)-5

$$x ( x - 1 ) - x ( 2 x + 1 ) = 2 ( 1 + x ) - 5$$

Solve for x

$$x=\sqrt{7}-2\approx 0.645751311$$

$$x=-\left(\sqrt{7}+2\right)\approx -4.645751311$$

Steps Using the Quadratic Formula

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

$$x^{2}-x-x\left(2x+1\right)=2\left(1+x\right)-5$$

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

$$x^{2}-x-\left(2x^{2}+x\right)=2\left(1+x\right)-5$$

To find the opposite of 2x^{2}+x, find the opposite of each term.

$$x^{2}-x-2x^{2}-x=2\left(1+x\right)-5$$

Combine x^{2} and -2x^{2} to get -x^{2}.

$$-x^{2}-x-x=2\left(1+x\right)-5$$

Combine -x and -x to get -2x.

$$-x^{2}-2x=2\left(1+x\right)-5$$

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

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

Subtract 5 from 2 to get -3.

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

Subtract -3 from both sides.

$$-x^{2}-2x-\left(-3\right)=2x$$

The opposite of -3 is 3.

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

Subtract 2x from both sides.

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

Combine -2x and -2x to get -4x.

$$-x^{2}-4x+3=0$$

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

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\times 3}}{2\left(-1\right)}$$

Multiply -4 times -1.

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

Multiply 4 times 3.

$$x=\frac{-\left(-4\right)±\sqrt{16+12}}{2\left(-1\right)}$$

Add 16 to 12.

$$x=\frac{-\left(-4\right)±\sqrt{28}}{2\left(-1\right)}$$

Take the square root of 28.

$$x=\frac{-\left(-4\right)±2\sqrt{7}}{2\left(-1\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{7}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{4±2\sqrt{7}}{-2}$$

Now solve the equation x=\frac{4±2\sqrt{7}}{-2} when ± is plus. Add 4 to 2\sqrt{7}.

$$x=\frac{2\sqrt{7}+4}{-2}$$

Divide 4+2\sqrt{7} by -2.

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

Now solve the equation x=\frac{4±2\sqrt{7}}{-2} when ± is minus. Subtract 2\sqrt{7} from 4.

$$x=\frac{4-2\sqrt{7}}{-2}$$

Divide 4-2\sqrt{7} by -2.

$$x=\sqrt{7}-2$$

The equation is now solved.

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

Steps for Completing the Square

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

$$x^{2}-x-x\left(2x+1\right)=2\left(1+x\right)-5$$

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

$$x^{2}-x-\left(2x^{2}+x\right)=2\left(1+x\right)-5$$

To find the opposite of 2x^{2}+x, find the opposite of each term.

$$x^{2}-x-2x^{2}-x=2\left(1+x\right)-5$$

Combine x^{2} and -2x^{2} to get -x^{2}.

$$-x^{2}-x-x=2\left(1+x\right)-5$$

Combine -x and -x to get -2x.

$$-x^{2}-2x=2\left(1+x\right)-5$$

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

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

Subtract 5 from 2 to get -3.

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

Subtract 2x from both sides.

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

Combine -2x and -2x to get -4x.

$$-x^{2}-4x=-3$$

Divide both sides by -1.

$$\frac{-x^{2}-4x}{-1}=-\frac{3}{-1}$$

Dividing by -1 undoes the multiplication by -1.

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

Divide -4 by -1.

$$x^{2}+4x=-\frac{3}{-1}$$

Divide -3 by -1.

$$x^{2}+4x=3$$

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

$$x^{2}+4x+2^{2}=3+2^{2}$$

Square 2.

$$x^{2}+4x+4=3+4$$

Add 3 to 4.

$$x^{2}+4x+4=7$$

Factor x^{2}+4x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+2=\sqrt{7} x+2=-\sqrt{7}$$

Subtract 2 from both sides of the equation.

$$x=\sqrt{7}-2 x=-\sqrt{7}-2$$