Solve 2x+3=x^2

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

Solve for x

$$x=-1$$

$$x=3$$

Steps Using Factoring By Grouping

Subtract x^{2} from both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=2 ab=-3=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=3 b=-1$$

Rewrite -x^{2}+2x+3 as \left(-x^{2}+3x\right)+\left(-x+3\right).

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

Factor out -x in the first and -1 in the second group.

$$-x\left(x-3\right)-\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(-x-1\right)$$

To find equation solutions, solve x-3=0 and -x-1=0.

$$x=3 x=-1$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

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

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.

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

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

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

Square 2.

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

Multiply -4 times -1.

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

Multiply 4 times 3.

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

Add 4 to 12.

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

Take the square root of 16.

$$x=\frac{-2±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-2±4}{-2}$$

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

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

Divide 2 by -2.

$$x=-1$$

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

$$x=-\frac{6}{-2}$$

Divide -6 by -2.

$$x=3$$

The equation is now solved.

$$x=-1 x=3$$

Steps for Completing the Square

Subtract x^{2} from both sides.

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

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

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 2 by -1.

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

Divide -3 by -1.

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

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

$$x^{2}-2x+1=3+1$$

Add 3 to 1.

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

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-1=2 x-1=-2$$

Add 1 to both sides of the equation.

$$x=3 x=-1$$