Solve 13x=x^2+30

$$13 x = x ^ { 2 } + 30$$

Solve for x

$$x=3$$

$$x=10$$

Steps Using Factoring By Grouping

Subtract x^{2} from both sides.

$$13x-x^{2}=30$$

Subtract 30 from both sides.

$$13x-x^{2}-30=0$$

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

$$-x^{2}+13x-30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-30. To find a and b, set up a system to be solved.

$$a+b=13 ab=-\left(-30\right)=30$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 30.

$$1,30 2,15 3,10 5,6$$

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 13.

$$a=10 b=3$$

Rewrite -x^{2}+13x-30 as \left(-x^{2}+10x\right)+\left(3x-30\right).

$$\left(-x^{2}+10x\right)+\left(3x-30\right)$$

Factor out -x in the first and 3 in the second group.

$$-x\left(x-10\right)+3\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(-x+3\right)$$

To find equation solutions, solve x-10=0 and -x+3=0.

$$x=10 x=3$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$13x-x^{2}=30$$

Subtract 30 from both sides.

$$13x-x^{2}-30=0$$

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.

$$-x^{2}+13x-30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 13 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-13±\sqrt{13^{2}-4\left(-1\right)\left(-30\right)}}{2\left(-1\right)}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\left(-1\right)\left(-30\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-13±\sqrt{169+4\left(-30\right)}}{2\left(-1\right)}$$

Multiply 4 times -30.

$$x=\frac{-13±\sqrt{169-120}}{2\left(-1\right)}$$

Add 169 to -120.

$$x=\frac{-13±\sqrt{49}}{2\left(-1\right)}$$

Take the square root of 49.

$$x=\frac{-13±7}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-13±7}{-2}$$

Now solve the equation x=\frac{-13±7}{-2} when ± is plus. Add -13 to 7.

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

Divide -6 by -2.

$$x=3$$

Now solve the equation x=\frac{-13±7}{-2} when ± is minus. Subtract 7 from -13.

$$x=-\frac{20}{-2}$$

Divide -20 by -2.

$$x=10$$

The equation is now solved.

$$x=3 x=10$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$13x-x^{2}=30$$

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.

$$-x^{2}+13x=30$$

Divide both sides by -1.

$$\frac{-x^{2}+13x}{-1}=\frac{30}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{13}{-1}x=\frac{30}{-1}$$

Divide 13 by -1.

$$x^{2}-13x=\frac{30}{-1}$$

Divide 30 by -1.

$$x^{2}-13x=-30$$

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.

$$x^{2}-13x+\left(-\frac{13}{2}\right)^{2}=-30+\left(-\frac{13}{2}\right)^{2}$$

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

$$x^{2}-13x+\frac{169}{4}=-30+\frac{169}{4}$$

Add -30 to \frac{169}{4}.

$$x^{2}-13x+\frac{169}{4}=\frac{49}{4}$$

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}.

$$\left(x-\frac{13}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x-\frac{13}{2}=\frac{7}{2} x-\frac{13}{2}=-\frac{7}{2}$$

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

$$x=10 x=3$$