Solve 9=49+x^2-13x

$$9 = 49 + x ^ { 2 } - 13 x$$

Solve for x

$$x=5$$

$$x=8$$

Steps Using Factoring

Swap sides so that all variable terms are on the left hand side.

$$49+x^{2}-13x=9$$

Subtract 9 from both sides.

$$49+x^{2}-13x-9=0$$

Subtract 9 from 49 to get 40.

$$40+x^{2}-13x=0$$

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

$$x^{2}-13x+40=0$$

To solve the equation, factor x^{2}-13x+40 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-13 ab=40$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 40.

$$-1,-40 -2,-20 -4,-10 -5,-8$$

Calculate the sum for each pair.

$$-1-40=-41 -2-20=-22 -4-10=-14 -5-8=-13$$

The solution is the pair that gives sum -13.

$$a=-8 b=-5$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-8\right)\left(x-5\right)$$

To find equation solutions, solve x-8=0 and x-5=0.

$$x=8 x=5$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$49+x^{2}-13x=9$$

Subtract 9 from both sides.

$$49+x^{2}-13x-9=0$$

Subtract 9 from 49 to get 40.

$$40+x^{2}-13x=0$$

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

$$x^{2}-13x+40=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+40. To find a and b, set up a system to be solved.

$$a+b=-13 ab=1\times 40=40$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 40.

$$-1,-40 -2,-20 -4,-10 -5,-8$$

Calculate the sum for each pair.

$$-1-40=-41 -2-20=-22 -4-10=-14 -5-8=-13$$

The solution is the pair that gives sum -13.

$$a=-8 b=-5$$

Rewrite x^{2}-13x+40 as \left(x^{2}-8x\right)+\left(-5x+40\right).

$$\left(x^{2}-8x\right)+\left(-5x+40\right)$$

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

$$x\left(x-8\right)-5\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(x-5\right)$$

To find equation solutions, solve x-8=0 and x-5=0.

$$x=8 x=5$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$49+x^{2}-13x=9$$

Subtract 9 from both sides.

$$49+x^{2}-13x-9=0$$

Subtract 9 from 49 to get 40.

$$40+x^{2}-13x=0$$

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.

$$x^{2}-13x+40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -13 for b, and 40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 40}}{2}$$

Square -13.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\times 40}}{2}$$

Multiply -4 times 40.

$$x=\frac{-\left(-13\right)±\sqrt{169-160}}{2}$$

Add 169 to -160.

$$x=\frac{-\left(-13\right)±\sqrt{9}}{2}$$

Take the square root of 9.

$$x=\frac{-\left(-13\right)±3}{2}$$

The opposite of -13 is 13.

$$x=\frac{13±3}{2}$$

Now solve the equation x=\frac{13±3}{2} when ± is plus. Add 13 to 3.

$$x=\frac{16}{2}$$

Divide 16 by 2.

$$x=8$$

Now solve the equation x=\frac{13±3}{2} when ± is minus. Subtract 3 from 13.

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

Divide 10 by 2.

$$x=5$$

The equation is now solved.

$$x=8 x=5$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$49+x^{2}-13x=9$$

Subtract 49 from both sides.

$$x^{2}-13x=9-49$$

Subtract 49 from 9 to get -40.

$$x^{2}-13x=-40$$

Divide -13, the coefficient of the x term, by 2 to get -\frac{13}{2}. Then add the square of -\frac{13}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-13x+\left(-\frac{13}{2}\right)^{2}=-40+\left(-\frac{13}{2}\right)^{2}$$

Square -\frac{13}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-13x+\frac{169}{4}=-40+\frac{169}{4}$$

Add -40 to \frac{169}{4}.

$$x^{2}-13x+\frac{169}{4}=\frac{9}{4}$$

Factor x^{2}-13x+\frac{169}{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}.

$$\left(x-\frac{13}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x-\frac{13}{2}=\frac{3}{2} x-\frac{13}{2}=-\frac{3}{2}$$

Add \frac{13}{2} to both sides of the equation.

$$x=8 x=5$$