Solve 2+x=left(x-2right)left(2x+3right)

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

Solve for x

$$x = \frac{\sqrt{17} + 1}{2} \approx 2.561552813$$

$$x=\frac{1-\sqrt{17}}{2}\approx -1.561552813$$

Steps Using the Quadratic Formula

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

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

Subtract 2x^{2} from both sides.

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

Add x to both sides.

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

Combine x and x to get 2x.

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

Add 6 to both sides.

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

Add 2 and 6 to get 8.

$$8+2x-2x^{2}=0$$

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.

$$-2x^{2}+2x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 2 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-2\right)\times 8}}{2\left(-2\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-2\right)\times 8}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-2±\sqrt{4+8\times 8}}{2\left(-2\right)}$$

Multiply 8 times 8.

$$x=\frac{-2±\sqrt{4+64}}{2\left(-2\right)}$$

Add 4 to 64.

$$x=\frac{-2±\sqrt{68}}{2\left(-2\right)}$$

Take the square root of 68.

$$x=\frac{-2±2\sqrt{17}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-2±2\sqrt{17}}{-4}$$

Now solve the equation x=\frac{-2±2\sqrt{17}}{-4} when ± is plus. Add -2 to 2\sqrt{17}.

$$x=\frac{2\sqrt{17}-2}{-4}$$

Divide -2+2\sqrt{17} by -4.

$$x=\frac{1-\sqrt{17}}{2}$$

Now solve the equation x=\frac{-2±2\sqrt{17}}{-4} when ± is minus. Subtract 2\sqrt{17} from -2.

$$x=\frac{-2\sqrt{17}-2}{-4}$$

Divide -2-2\sqrt{17} by -4.

$$x=\frac{\sqrt{17}+1}{2}$$

The equation is now solved.

$$x=\frac{1-\sqrt{17}}{2} x=\frac{\sqrt{17}+1}{2}$$

Steps for Completing the Square

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

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

Subtract 2x^{2} from both sides.

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

Add x to both sides.

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

Combine x and x to get 2x.

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

Subtract 2 from both sides.

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

Subtract 2 from -6 to get -8.

$$2x-2x^{2}=-8$$

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.

$$-2x^{2}+2x=-8$$

Divide both sides by -2.

$$\frac{-2x^{2}+2x}{-2}=-\frac{8}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{2}{-2}x=-\frac{8}{-2}$$

Divide 2 by -2.

$$x^{2}-x=-\frac{8}{-2}$$

Divide -8 by -2.

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

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add 4 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{17}{4}$$

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{17}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{17}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{17}}{2} x-\frac{1}{2}=-\frac{\sqrt{17}}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{\sqrt{17}+1}{2} x=\frac{1-\sqrt{17}}{2}$$