Solve x^2+20=9x

$$x ^ { 2 } + 20 = 9 x$$

Solve for x

$$x=4$$

$$x=5$$

Steps Using Factoring

Subtract 9x from both sides.

$$x^{2}+20-9x=0$$

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

$$x^{2}-9x+20=0$$

To solve the equation, factor x^{2}-9x+20 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.

$$a+b=-9 ab=20$$

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 20.

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

Calculate the sum for each pair.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

The solution is the pair that gives sum -9.

$$a=-5 b=-4$$

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

$$\left(x-5\right)\left(x-4\right)$$

To find equation solutions, solve x-5=0 and x-4=0.

$$x=5 x=4$$

Steps Using Factoring By Grouping

Subtract 9x from both sides.

$$x^{2}+20-9x=0$$

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

$$x^{2}-9x+20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+20. To find a and b, set up a system to be solved.

$$a+b=-9 ab=1\times 20=20$$

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 20.

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

Calculate the sum for each pair.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

The solution is the pair that gives sum -9.

$$a=-5 b=-4$$

Rewrite x^{2}-9x+20 as \left(x^{2}-5x\right)+\left(-4x+20\right).

$$\left(x^{2}-5x\right)+\left(-4x+20\right)$$

Factor out x in the first and -4 in the second group.

$$x\left(x-5\right)-4\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x-4\right)$$

To find equation solutions, solve x-5=0 and x-4=0.

$$x=5 x=4$$

Steps Using the Quadratic Formula

Subtract 9x from both sides.

$$x^{2}+20-9x=0$$

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.

$$x^{2}-9x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -9 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 20}}{2}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 20}}{2}$$

Multiply -4 times 20.

$$x=\frac{-\left(-9\right)±\sqrt{81-80}}{2}$$

Add 81 to -80.

$$x=\frac{-\left(-9\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$x=\frac{-\left(-9\right)±1}{2}$$

The opposite of -9 is 9.

$$x=\frac{9±1}{2}$$

Now solve the equation x=\frac{9±1}{2} when ± is plus. Add 9 to 1.

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

Divide 10 by 2.

$$x=5$$

Now solve the equation x=\frac{9±1}{2} when ± is minus. Subtract 1 from 9.

$$x=\frac{8}{2}$$

Divide 8 by 2.

$$x=4$$

The equation is now solved.

$$x=5 x=4$$

Steps for Completing the Square

Subtract 9x from both sides.

$$x^{2}+20-9x=0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-9x=-20$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=-20+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=-20+\frac{81}{4}$$

Add -20 to \frac{81}{4}.

$$x^{2}-9x+\frac{81}{4}=\frac{1}{4}$$

Factor x^{2}-9x+\frac{81}{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}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{1}{2} x-\frac{9}{2}=-\frac{1}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=5 x=4$$