Solve (x^2-9x+20)=(x+2)=

$$( x ^ { 2 } - 9 x + 20 ) = ( x + 2 ) =$$

Solve for x

$$x=\sqrt{7}+5\approx 7.645751311$$

$$x=5-\sqrt{7}\approx 2.354248689$$

Steps Using the Quadratic Formula

Subtract x from both sides.

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

Combine -9x and -x to get -10x.

$$x^{2}-10x+20=2$$

Subtract 2 from both sides.

$$x^{2}-10x+20-2=0$$

Subtract 2 from 20 to get 18.

$$x^{2}-10x+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 18}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 18}}{2}$$

Multiply -4 times 18.

$$x=\frac{-\left(-10\right)±\sqrt{100-72}}{2}$$

Add 100 to -72.

$$x=\frac{-\left(-10\right)±\sqrt{28}}{2}$$

Take the square root of 28.

$$x=\frac{-\left(-10\right)±2\sqrt{7}}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{7}}{2}$$

Now solve the equation x=\frac{10±2\sqrt{7}}{2} when ± is plus. Add 10 to 2\sqrt{7}.

$$x=\frac{2\sqrt{7}+10}{2}$$

Divide 10+2\sqrt{7} by 2.

$$x=\sqrt{7}+5$$

Now solve the equation x=\frac{10±2\sqrt{7}}{2} when ± is minus. Subtract 2\sqrt{7} from 10.

$$x=\frac{10-2\sqrt{7}}{2}$$

Divide 10-2\sqrt{7} by 2.

$$x=5-\sqrt{7}$$

The equation is now solved.

$$x=\sqrt{7}+5 x=5-\sqrt{7}$$

Steps for Completing the Square

Subtract x from both sides.

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

Combine -9x and -x to get -10x.

$$x^{2}-10x+20=2$$

Subtract 20 from both sides.

$$x^{2}-10x=2-20$$

Subtract 20 from 2 to get -18.

$$x^{2}-10x=-18$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-18+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-18+25$$

Add -18 to 25.

$$x^{2}-10x+25=7$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=7$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{7}$$

Simplify.

$$x-5=\sqrt{7} x-5=-\sqrt{7}$$

Add 5 to both sides of the equation.

$$x=\sqrt{7}+5 x=5-\sqrt{7}$$