Solve (90-30x)^2+(20x)^2=50^2

$$( 90 - 30 x ) ^ { 2 } + ( 20 x ) ^ { 2 } = 50 ^ { 2 }$$

Solve for x

$$x=2$$

$$x = \frac{28}{13} = 2\frac{2}{13} \approx 2.153846154$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(90-30x\right)^{2}.

$$8100-5400x+900x^{2}+\left(20x\right)^{2}=50^{2}$$

Expand \left(20x\right)^{2}.

$$8100-5400x+900x^{2}+20^{2}x^{2}=50^{2}$$

Calculate 20 to the power of 2 and get 400.

$$8100-5400x+900x^{2}+400x^{2}=50^{2}$$

Combine 900x^{2} and 400x^{2} to get 1300x^{2}.

$$8100-5400x+1300x^{2}=50^{2}$$

Calculate 50 to the power of 2 and get 2500.

$$8100-5400x+1300x^{2}=2500$$

Subtract 2500 from both sides.

$$8100-5400x+1300x^{2}-2500=0$$

Subtract 2500 from 8100 to get 5600.

$$5600-5400x+1300x^{2}=0$$

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.

$$1300x^{2}-5400x+5600=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1300 for a, -5400 for b, and 5600 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5400\right)±\sqrt{\left(-5400\right)^{2}-4\times 1300\times 5600}}{2\times 1300}$$

Square -5400.

$$x=\frac{-\left(-5400\right)±\sqrt{29160000-4\times 1300\times 5600}}{2\times 1300}$$

Multiply -4 times 1300.

$$x=\frac{-\left(-5400\right)±\sqrt{29160000-5200\times 5600}}{2\times 1300}$$

Multiply -5200 times 5600.

$$x=\frac{-\left(-5400\right)±\sqrt{29160000-29120000}}{2\times 1300}$$

Add 29160000 to -29120000.

$$x=\frac{-\left(-5400\right)±\sqrt{40000}}{2\times 1300}$$

Take the square root of 40000.

$$x=\frac{-\left(-5400\right)±200}{2\times 1300}$$

The opposite of -5400 is 5400.

$$x=\frac{5400±200}{2\times 1300}$$

Multiply 2 times 1300.

$$x=\frac{5400±200}{2600}$$

Now solve the equation x=\frac{5400±200}{2600} when ± is plus. Add 5400 to 200.

$$x=\frac{5600}{2600}$$

Reduce the fraction \frac{5600}{2600} to lowest terms by extracting and canceling out 200.

$$x=\frac{28}{13}$$

Now solve the equation x=\frac{5400±200}{2600} when ± is minus. Subtract 200 from 5400.

$$x=\frac{5200}{2600}$$

Divide 5200 by 2600.

$$x=2$$

The equation is now solved.

$$x=\frac{28}{13} x=2$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(90-30x\right)^{2}.

$$8100-5400x+900x^{2}+\left(20x\right)^{2}=50^{2}$$

Expand \left(20x\right)^{2}.

$$8100-5400x+900x^{2}+20^{2}x^{2}=50^{2}$$

Calculate 20 to the power of 2 and get 400.

$$8100-5400x+900x^{2}+400x^{2}=50^{2}$$

Combine 900x^{2} and 400x^{2} to get 1300x^{2}.

$$8100-5400x+1300x^{2}=50^{2}$$

Calculate 50 to the power of 2 and get 2500.

$$8100-5400x+1300x^{2}=2500$$

Subtract 8100 from both sides.

$$-5400x+1300x^{2}=2500-8100$$

Subtract 8100 from 2500 to get -5600.

$$-5400x+1300x^{2}=-5600$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$1300x^{2}-5400x=-5600$$

Divide both sides by 1300.

$$\frac{1300x^{2}-5400x}{1300}=-\frac{5600}{1300}$$

Dividing by 1300 undoes the multiplication by 1300.

$$x^{2}+\left(-\frac{5400}{1300}\right)x=-\frac{5600}{1300}$$

Reduce the fraction \frac{-5400}{1300} to lowest terms by extracting and canceling out 100.

$$x^{2}-\frac{54}{13}x=-\frac{5600}{1300}$$

Reduce the fraction \frac{-5600}{1300} to lowest terms by extracting and canceling out 100.

$$x^{2}-\frac{54}{13}x=-\frac{56}{13}$$

Divide -\frac{54}{13}, the coefficient of the x term, by 2 to get -\frac{27}{13}. Then add the square of -\frac{27}{13} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{54}{13}x+\left(-\frac{27}{13}\right)^{2}=-\frac{56}{13}+\left(-\frac{27}{13}\right)^{2}$$

Square -\frac{27}{13} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{54}{13}x+\frac{729}{169}=-\frac{56}{13}+\frac{729}{169}$$

Add -\frac{56}{13} to \frac{729}{169} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{54}{13}x+\frac{729}{169}=\frac{1}{169}$$

Factor x^{2}-\frac{54}{13}x+\frac{729}{169}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{27}{13}\right)^{2}=\frac{1}{169}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{27}{13}\right)^{2}}=\sqrt{\frac{1}{169}}$$

Simplify.

$$x-\frac{27}{13}=\frac{1}{13} x-\frac{27}{13}=-\frac{1}{13}$$

Add \frac{27}{13} to both sides of the equation.

$$x=\frac{28}{13} x=2$$