Solve 4x+45=x^2

$$4x+45= { x }^{ 2 }$$

Solve for x

$$x=-5$$

$$x=9$$

Steps Using Factoring By Grouping

Subtract x^{2} from both sides.

$$4x+45-x^{2}=0$$

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

$$-x^{2}+4x+45=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+45. To find a and b, set up a system to be solved.

$$a+b=4 ab=-45=-45$$

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. List all such integer pairs that give product -45.

$$-1,45 -3,15 -5,9$$

Calculate the sum for each pair.

$$-1+45=44 -3+15=12 -5+9=4$$

The solution is the pair that gives sum 4.

$$a=9 b=-5$$

Rewrite -x^{2}+4x+45 as \left(-x^{2}+9x\right)+\left(-5x+45\right).

$$\left(-x^{2}+9x\right)+\left(-5x+45\right)$$

Factor out -x in the first and -5 in the second group.

$$-x\left(x-9\right)-5\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(-x-5\right)$$

To find equation solutions, solve x-9=0 and -x-5=0.

$$x=9 x=-5$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$4x+45-x^{2}=0$$

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.

$$-x^{2}+4x+45=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 4 for b, and 45 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-1\right)\times 45}}{2\left(-1\right)}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-1\right)\times 45}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-4±\sqrt{16+4\times 45}}{2\left(-1\right)}$$

Multiply 4 times 45.

$$x=\frac{-4±\sqrt{16+180}}{2\left(-1\right)}$$

Add 16 to 180.

$$x=\frac{-4±\sqrt{196}}{2\left(-1\right)}$$

Take the square root of 196.

$$x=\frac{-4±14}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-4±14}{-2}$$

Now solve the equation x=\frac{-4±14}{-2} when ± is plus. Add -4 to 14.

$$x=\frac{10}{-2}$$

Divide 10 by -2.

$$x=-5$$

Now solve the equation x=\frac{-4±14}{-2} when ± is minus. Subtract 14 from -4.

$$x=-\frac{18}{-2}$$

Divide -18 by -2.

$$x=9$$

The equation is now solved.

$$x=-5 x=9$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$4x+45-x^{2}=0$$

Subtract 45 from both sides. Anything subtracted from zero gives its negation.

$$4x-x^{2}=-45$$

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.

$$-x^{2}+4x=-45$$

Divide both sides by -1.

$$\frac{-x^{2}+4x}{-1}=-\frac{45}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{4}{-1}x=-\frac{45}{-1}$$

Divide 4 by -1.

$$x^{2}-4x=-\frac{45}{-1}$$

Divide -45 by -1.

$$x^{2}-4x=45$$

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.

$$x^{2}-4x+\left(-2\right)^{2}=45+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=45+4$$

Add 45 to 4.

$$x^{2}-4x+4=49$$

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}.

$$\left(x-2\right)^{2}=49$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{49}$$

Simplify.

$$x-2=7 x-2=-7$$

Add 2 to both sides of the equation.

$$x=9 x=-5$$