Solve x^2+(40-4x)^2=58

$$x ^ { 2 } + ( 40 - 4 x ) ^ { 2 } = 58$$

Solve for x (complex solution)

$$x=\frac{160+\sqrt{614}i}{17}\approx 9.411764706+1.457589611i$$

$$x=\frac{-\sqrt{614}i+160}{17}\approx 9.411764706-1.457589611i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(40-4x\right)^{2}.

$$x^{2}+1600-320x+16x^{2}=58$$

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

$$17x^{2}+1600-320x=58$$

Subtract 58 from both sides.

$$17x^{2}+1600-320x-58=0$$

Subtract 58 from 1600 to get 1542.

$$17x^{2}+1542-320x=0$$

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.

$$17x^{2}-320x+1542=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 17 for a, -320 for b, and 1542 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-320\right)±\sqrt{\left(-320\right)^{2}-4\times 17\times 1542}}{2\times 17}$$

Square -320.

$$x=\frac{-\left(-320\right)±\sqrt{102400-4\times 17\times 1542}}{2\times 17}$$

Multiply -4 times 17.

$$x=\frac{-\left(-320\right)±\sqrt{102400-68\times 1542}}{2\times 17}$$

Multiply -68 times 1542.

$$x=\frac{-\left(-320\right)±\sqrt{102400-104856}}{2\times 17}$$

Add 102400 to -104856.

$$x=\frac{-\left(-320\right)±\sqrt{-2456}}{2\times 17}$$

Take the square root of -2456.

$$x=\frac{-\left(-320\right)±2\sqrt{614}i}{2\times 17}$$

The opposite of -320 is 320.

$$x=\frac{320±2\sqrt{614}i}{2\times 17}$$

Multiply 2 times 17.

$$x=\frac{320±2\sqrt{614}i}{34}$$

Now solve the equation x=\frac{320±2\sqrt{614}i}{34} when ± is plus. Add 320 to 2i\sqrt{614}.

$$x=\frac{320+2\sqrt{614}i}{34}$$

Divide 320+2i\sqrt{614} by 34.

$$x=\frac{160+\sqrt{614}i}{17}$$

Now solve the equation x=\frac{320±2\sqrt{614}i}{34} when ± is minus. Subtract 2i\sqrt{614} from 320.

$$x=\frac{-2\sqrt{614}i+320}{34}$$

Divide 320-2i\sqrt{614} by 34.

$$x=\frac{-\sqrt{614}i+160}{17}$$

The equation is now solved.

$$x=\frac{160+\sqrt{614}i}{17} x=\frac{-\sqrt{614}i+160}{17}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(40-4x\right)^{2}.

$$x^{2}+1600-320x+16x^{2}=58$$

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

$$17x^{2}+1600-320x=58$$

Subtract 1600 from both sides.

$$17x^{2}-320x=58-1600$$

Subtract 1600 from 58 to get -1542.

$$17x^{2}-320x=-1542$$

Divide both sides by 17.

$$\frac{17x^{2}-320x}{17}=-\frac{1542}{17}$$

Dividing by 17 undoes the multiplication by 17.

$$x^{2}-\frac{320}{17}x=-\frac{1542}{17}$$

Divide -\frac{320}{17}, the coefficient of the x term, by 2 to get -\frac{160}{17}. Then add the square of -\frac{160}{17} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{320}{17}x+\left(-\frac{160}{17}\right)^{2}=-\frac{1542}{17}+\left(-\frac{160}{17}\right)^{2}$$

Square -\frac{160}{17} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{320}{17}x+\frac{25600}{289}=-\frac{1542}{17}+\frac{25600}{289}$$

Add -\frac{1542}{17} to \frac{25600}{289} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{320}{17}x+\frac{25600}{289}=-\frac{614}{289}$$

Factor x^{2}-\frac{320}{17}x+\frac{25600}{289}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{160}{17}\right)^{2}=-\frac{614}{289}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{160}{17}\right)^{2}}=\sqrt{-\frac{614}{289}}$$

Simplify.

$$x-\frac{160}{17}=\frac{\sqrt{614}i}{17} x-\frac{160}{17}=-\frac{\sqrt{614}i}{17}$$

Add \frac{160}{17} to both sides of the equation.

$$x=\frac{160+\sqrt{614}i}{17} x=\frac{-\sqrt{614}i+160}{17}$$