Solve x+x*(x-3)+3=x*(4-2x)+x

$$x + x \cdot ( x - 3 ) + 3 = x \cdot ( 4 - 2 x ) + x$$

Solve for x

$$x = \frac{\sqrt{13} + 7}{6} \approx 1.767591879$$

$$x=\frac{7-\sqrt{13}}{6}\approx 0.565741454$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x by x-3.

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

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

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

Use the distributive property to multiply x by 4-2x.

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

Combine 4x and x to get 5x.

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

Subtract 5x from both sides.

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

Combine -2x and -5x to get -7x.

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

Add 2x^{2} to both sides.

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

Combine x^{2} and 2x^{2} to get 3x^{2}.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -7 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 3\times 3}}{2\times 3}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 3\times 3}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-7\right)±\sqrt{49-12\times 3}}{2\times 3}$$

Multiply -12 times 3.

$$x=\frac{-\left(-7\right)±\sqrt{49-36}}{2\times 3}$$

Add 49 to -36.

$$x=\frac{-\left(-7\right)±\sqrt{13}}{2\times 3}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{13}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{7±\sqrt{13}}{6}$$

Now solve the equation x=\frac{7±\sqrt{13}}{6} when ± is plus. Add 7 to \sqrt{13}.

$$x=\frac{\sqrt{13}+7}{6}$$

Now solve the equation x=\frac{7±\sqrt{13}}{6} when ± is minus. Subtract \sqrt{13} from 7.

$$x=\frac{7-\sqrt{13}}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{13}+7}{6} x=\frac{7-\sqrt{13}}{6}$$

Steps for Completing the Square

Use the distributive property to multiply x by x-3.

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

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

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

Use the distributive property to multiply x by 4-2x.

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

Combine 4x and x to get 5x.

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

Subtract 5x from both sides.

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

Combine -2x and -5x to get -7x.

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

Add 2x^{2} to both sides.

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

Combine x^{2} and 2x^{2} to get 3x^{2}.

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

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Divide both sides by 3.

$$\frac{3x^{2}-7x}{3}=-\frac{3}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{7}{3}x=-\frac{3}{3}$$

Divide -3 by 3.

$$x^{2}-\frac{7}{3}x=-1$$

Divide -\frac{7}{3}, the coefficient of the x term, by 2 to get -\frac{7}{6}. Then add the square of -\frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{3}x+\left(-\frac{7}{6}\right)^{2}=-1+\left(-\frac{7}{6}\right)^{2}$$

Square -\frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{3}x+\frac{49}{36}=-1+\frac{49}{36}$$

Add -1 to \frac{49}{36}.

$$x^{2}-\frac{7}{3}x+\frac{49}{36}=\frac{13}{36}$$

Factor x^{2}-\frac{7}{3}x+\frac{49}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{6}\right)^{2}=\frac{13}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{6}\right)^{2}}=\sqrt{\frac{13}{36}}$$

Simplify.

$$x-\frac{7}{6}=\frac{\sqrt{13}}{6} x-\frac{7}{6}=-\frac{\sqrt{13}}{6}$$

Add \frac{7}{6} to both sides of the equation.

$$x=\frac{\sqrt{13}+7}{6} x=\frac{7-\sqrt{13}}{6}$$