Solve (x-5)^2+(4x-22+2)^2=34

$$( x - 5 ) ^ { 2 } + ( 4 x - 22 + 2 ) ^ { 2 } = 34$$

Solve for x

$$x=\sqrt{2}+5\approx 6.414213562$$

$$x=5-\sqrt{2}\approx 3.585786438$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}-10x+25+\left(4x-22+2\right)^{2}=34$$

Add -22 and 2 to get -20.

$$x^{2}-10x+25+\left(4x-20\right)^{2}=34$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-20\right)^{2}.

$$x^{2}-10x+25+16x^{2}-160x+400=34$$

Combine x^{2} and 16x^{2} to get 17x^{2}.

$$17x^{2}-10x+25-160x+400=34$$

Combine -10x and -160x to get -170x.

$$17x^{2}-170x+25+400=34$$

Add 25 and 400 to get 425.

$$17x^{2}-170x+425=34$$

Subtract 34 from both sides.

$$17x^{2}-170x+425-34=0$$

Subtract 34 from 425 to get 391.

$$17x^{2}-170x+391=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 17 for a, -170 for b, and 391 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-170\right)±\sqrt{\left(-170\right)^{2}-4\times 17\times 391}}{2\times 17}$$

Square -170.

$$x=\frac{-\left(-170\right)±\sqrt{28900-4\times 17\times 391}}{2\times 17}$$

Multiply -4 times 17.

$$x=\frac{-\left(-170\right)±\sqrt{28900-68\times 391}}{2\times 17}$$

Multiply -68 times 391.

$$x=\frac{-\left(-170\right)±\sqrt{28900-26588}}{2\times 17}$$

Add 28900 to -26588.

$$x=\frac{-\left(-170\right)±\sqrt{2312}}{2\times 17}$$

Take the square root of 2312.

$$x=\frac{-\left(-170\right)±34\sqrt{2}}{2\times 17}$$

The opposite of -170 is 170.

$$x=\frac{170±34\sqrt{2}}{2\times 17}$$

Multiply 2 times 17.

$$x=\frac{170±34\sqrt{2}}{34}$$

Now solve the equation x=\frac{170±34\sqrt{2}}{34} when ± is plus. Add 170 to 34\sqrt{2}.

$$x=\frac{34\sqrt{2}+170}{34}$$

Divide 170+34\sqrt{2} by 34.

$$x=\sqrt{2}+5$$

Now solve the equation x=\frac{170±34\sqrt{2}}{34} when ± is minus. Subtract 34\sqrt{2} from 170.

$$x=\frac{170-34\sqrt{2}}{34}$$

Divide 170-34\sqrt{2} by 34.

$$x=5-\sqrt{2}$$

The equation is now solved.

$$x=\sqrt{2}+5 x=5-\sqrt{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}-10x+25+\left(4x-22+2\right)^{2}=34$$

Add -22 and 2 to get -20.

$$x^{2}-10x+25+\left(4x-20\right)^{2}=34$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-20\right)^{2}.

$$x^{2}-10x+25+16x^{2}-160x+400=34$$

Combine x^{2} and 16x^{2} to get 17x^{2}.

$$17x^{2}-10x+25-160x+400=34$$

Combine -10x and -160x to get -170x.

$$17x^{2}-170x+25+400=34$$

Add 25 and 400 to get 425.

$$17x^{2}-170x+425=34$$

Subtract 425 from both sides.

$$17x^{2}-170x=34-425$$

Subtract 425 from 34 to get -391.

$$17x^{2}-170x=-391$$

Divide both sides by 17.

$$\frac{17x^{2}-170x}{17}=-\frac{391}{17}$$

Dividing by 17 undoes the multiplication by 17.

$$x^{2}+\left(-\frac{170}{17}\right)x=-\frac{391}{17}$$

Divide -170 by 17.

$$x^{2}-10x=-\frac{391}{17}$$

Divide -391 by 17.

$$x^{2}-10x=-23$$

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.

$$x^{2}-10x+\left(-5\right)^{2}=-23+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-23+25$$

Add -23 to 25.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-5=\sqrt{2} x-5=-\sqrt{2}$$

Add 5 to both sides of the equation.

$$x=\sqrt{2}+5 x=5-\sqrt{2}$$