Solve 11x=x^2+30

$$11x= { x }^{ 2 } +30$$

Solve for x

$$x=5$$

$$x=6$$

Steps Using Factoring By Grouping

Subtract x^{2} from both sides.

$$11x-x^{2}=30$$

Subtract 30 from both sides.

$$11x-x^{2}-30=0$$

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

$$-x^{2}+11x-30=0$$

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.

$$a+b=11 ab=-\left(-30\right)=30$$

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 11.

$$a=6 b=5$$

Rewrite -x^{2}+11x-30 as \left(-x^{2}+6x\right)+\left(5x-30\right).

$$\left(-x^{2}+6x\right)+\left(5x-30\right)$$

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

$$-x\left(x-6\right)+5\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(-x+5\right)$$

To find equation solutions, solve x-6=0 and -x+5=0.

$$x=6 x=5$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$11x-x^{2}=30$$

Subtract 30 from both sides.

$$11x-x^{2}-30=0$$

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.

$$-x^{2}+11x-30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 11 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-1\right)\left(-30\right)}}{2\left(-1\right)}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\left(-1\right)\left(-30\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-11±\sqrt{121+4\left(-30\right)}}{2\left(-1\right)}$$

Multiply 4 times -30.

$$x=\frac{-11±\sqrt{121-120}}{2\left(-1\right)}$$

Add 121 to -120.

$$x=\frac{-11±\sqrt{1}}{2\left(-1\right)}$$

Take the square root of 1.

$$x=\frac{-11±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-11±1}{-2}$$

Now solve the equation x=\frac{-11±1}{-2} when ± is plus. Add -11 to 1.

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

Divide -10 by -2.

$$x=5$$

Now solve the equation x=\frac{-11±1}{-2} when ± is minus. Subtract 1 from -11.

$$x=-\frac{12}{-2}$$

Divide -12 by -2.

$$x=6$$

The equation is now solved.

$$x=5 x=6$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$11x-x^{2}=30$$

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.

$$-x^{2}+11x=30$$

Divide both sides by -1.

$$\frac{-x^{2}+11x}{-1}=\frac{30}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{11}{-1}x=\frac{30}{-1}$$

Divide 11 by -1.

$$x^{2}-11x=\frac{30}{-1}$$

Divide 30 by -1.

$$x^{2}-11x=-30$$

Divide -11, the coefficient of the x term, by 2 to get -\frac{11}{2}. Then add the square of -\frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-11x+\left(-\frac{11}{2}\right)^{2}=-30+\left(-\frac{11}{2}\right)^{2}$$

Square -\frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-11x+\frac{121}{4}=-30+\frac{121}{4}$$

Add -30 to \frac{121}{4}.

$$x^{2}-11x+\frac{121}{4}=\frac{1}{4}$$

Factor x^{2}-11x+\frac{121}{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}.

$$\left(x-\frac{11}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{11}{2}=\frac{1}{2} x-\frac{11}{2}=-\frac{1}{2}$$

Add \frac{11}{2} to both sides of the equation.

$$x=6 x=5$$