Solve 3x-6x+3-2x(x+1)=3

$$3 x - 6 x + 3 - 2 x ( x + 1 ) = 3$$

Solve for x

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

$$x=0$$

Steps Using the Quadratic Formula

Combine 3x and -6x to get -3x.

$$-3x+3-2x\left(x+1\right)=3$$

Subtract 3 from both sides.

$$-3x+3-2x\left(x+1\right)-3=0$$

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

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

Combine -3x and -2x to get -5x.

$$-5x+3-2x^{2}-3=0$$

Subtract 3 from 3 to get 0.

$$-5x-2x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-2x^{2}-5x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -5 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}}}{2\left(-2\right)}$$

Take the square root of \left(-5\right)^{2}.

$$x=\frac{-\left(-5\right)±5}{2\left(-2\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±5}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{5±5}{-4}$$

Now solve the equation x=\frac{5±5}{-4} when ± is plus. Add 5 to 5.

$$x=\frac{10}{-4}$$

Reduce the fraction \frac{10}{-4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{5±5}{-4} when ± is minus. Subtract 5 from 5.

$$x=\frac{0}{-4}$$

Divide 0 by -4.

$$x=0$$

The equation is now solved.

$$x=-\frac{5}{2} x=0$$

Steps for Completing the Square

Combine 3x and -6x to get -3x.

$$-3x+3-2x\left(x+1\right)=3$$

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

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

Combine -3x and -2x to get -5x.

$$-5x+3-2x^{2}=3$$

Subtract 3 from both sides.

$$-5x-2x^{2}=3-3$$

Subtract 3 from 3 to get 0.

$$-5x-2x^{2}=0$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-2x^{2}-5x=0$$

Divide both sides by -2.

$$\frac{-2x^{2}-5x}{-2}=\frac{0}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{5}{-2}\right)x=\frac{0}{-2}$$

Divide -5 by -2.

$$x^{2}+\frac{5}{2}x=\frac{0}{-2}$$

Divide 0 by -2.

$$x^{2}+\frac{5}{2}x=0$$

Divide \frac{5}{2}, the coefficient of the x term, by 2 to get \frac{5}{4}. Then add the square of \frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{2}x+\left(\frac{5}{4}\right)^{2}=\left(\frac{5}{4}\right)^{2}$$

Square \frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{25}{16}$$

Factor x^{2}+\frac{5}{2}x+\frac{25}{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}.

$$\left(x+\frac{5}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x+\frac{5}{4}=\frac{5}{4} x+\frac{5}{4}=-\frac{5}{4}$$

Subtract \frac{5}{4} from both sides of the equation.

$$x=0 x=-\frac{5}{2}$$