Solve x^2+2x+4=22x+9

$${ x }^{ 2 } +2x+4=22x+9$$

Solve for x

$$x=\sqrt{105}+10\approx 20.246950766$$

$$x=10-\sqrt{105}\approx -0.246950766$$

Steps Using the Quadratic Formula

Subtract 22x from both sides.

$$x^{2}+2x+4-22x=9$$

Combine 2x and -22x to get -20x.

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

Subtract 9 from both sides.

$$x^{2}-20x+4-9=0$$

Subtract 9 from 4 to get -5.

$$x^{2}-20x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -20 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\left(-5\right)}}{2}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\left(-5\right)}}{2}$$

Multiply -4 times -5.

$$x=\frac{-\left(-20\right)±\sqrt{400+20}}{2}$$

Add 400 to 20.

$$x=\frac{-\left(-20\right)±\sqrt{420}}{2}$$

Take the square root of 420.

$$x=\frac{-\left(-20\right)±2\sqrt{105}}{2}$$

The opposite of -20 is 20.

$$x=\frac{20±2\sqrt{105}}{2}$$

Now solve the equation x=\frac{20±2\sqrt{105}}{2} when ± is plus. Add 20 to 2\sqrt{105}.

$$x=\frac{2\sqrt{105}+20}{2}$$

Divide 20+2\sqrt{105} by 2.

$$x=\sqrt{105}+10$$

Now solve the equation x=\frac{20±2\sqrt{105}}{2} when ± is minus. Subtract 2\sqrt{105} from 20.

$$x=\frac{20-2\sqrt{105}}{2}$$

Divide 20-2\sqrt{105} by 2.

$$x=10-\sqrt{105}$$

The equation is now solved.

$$x=\sqrt{105}+10 x=10-\sqrt{105}$$

Steps for Completing the Square

Subtract 22x from both sides.

$$x^{2}+2x+4-22x=9$$

Combine 2x and -22x to get -20x.

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

Subtract 4 from both sides.

$$x^{2}-20x=9-4$$

Subtract 4 from 9 to get 5.

$$x^{2}-20x=5$$

Divide -20, the coefficient of the x term, by 2 to get -10. Then add the square of -10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-20x+\left(-10\right)^{2}=5+\left(-10\right)^{2}$$

Square -10.

$$x^{2}-20x+100=5+100$$

Add 5 to 100.

$$x^{2}-20x+100=105$$

Factor x^{2}-20x+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-10\right)^{2}=105$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-10\right)^{2}}=\sqrt{105}$$

Simplify.

$$x-10=\sqrt{105} x-10=-\sqrt{105}$$

Add 10 to both sides of the equation.

$$x=\sqrt{105}+10 x=10-\sqrt{105}$$