Solve (20x)^2+5200+(60x)^2/40x*20sqrt{13}=frac{5sqrt{3}}{26}

$$\frac { ( 20 x ) ^ { 2 } + 5200 + ( 60 x ) ^ { 2 } } { 40 x \times 20 \sqrt { 13 } } = \frac { 5 \sqrt { 3 } } { 26 }$$

Solve for x (complex solution)

$$x=\frac{\sqrt{86905}i}{260}+\frac{\sqrt{39}}{52}\approx 0.120096115+1.133832846i$$

$$x=-\frac{\sqrt{86905}i}{260}+\frac{\sqrt{39}}{52}\approx 0.120096115-1.133832846i$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 26.

$$\frac{1}{400}x^{-1}\times 13^{\frac{1}{2}}\left(\left(20x\right)^{2}+5200+\left(60x\right)^{2}\right)=5\sqrt{3}$$

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

$$\frac{1}{400}x^{-1}\times 13^{\frac{1}{2}}\left(20^{2}x^{2}+5200+\left(60x\right)^{2}\right)=5\sqrt{3}$$

Calculate 20 to the power of 2 and get 400.

$$\frac{1}{400}x^{-1}\times 13^{\frac{1}{2}}\left(400x^{2}+5200+\left(60x\right)^{2}\right)=5\sqrt{3}$$

Expand \left(60x\right)^{2}.

$$\frac{1}{400}x^{-1}\times 13^{\frac{1}{2}}\left(400x^{2}+5200+60^{2}x^{2}\right)=5\sqrt{3}$$

Calculate 60 to the power of 2 and get 3600.

$$\frac{1}{400}x^{-1}\times 13^{\frac{1}{2}}\left(400x^{2}+5200+3600x^{2}\right)=5\sqrt{3}$$

Combine 400x^{2} and 3600x^{2} to get 4000x^{2}.

$$\frac{1}{400}x^{-1}\times 13^{\frac{1}{2}}\left(4000x^{2}+5200\right)=5\sqrt{3}$$

Use the distributive property to multiply \frac{1}{400}x^{-1}\times 13^{\frac{1}{2}} by 4000x^{2}+5200.

$$10x\times 13^{\frac{1}{2}}+13x^{-1}\times 13^{\frac{1}{2}}=5\sqrt{3}$$

To multiply powers of the same base, add their exponents. Add 1 and \frac{1}{2} to get \frac{3}{2}.

$$10x\times 13^{\frac{1}{2}}+13^{\frac{3}{2}}x^{-1}=5\sqrt{3}$$

Subtract 5\sqrt{3} from both sides.

$$10x\times 13^{\frac{1}{2}}+13^{\frac{3}{2}}x^{-1}-5\sqrt{3}=0$$

Reorder the terms.

$$10\sqrt{13}x-5\sqrt{3}+13^{\frac{3}{2}}\times \frac{1}{x}=0$$

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$10\sqrt{13}xx-5\sqrt{3}x+13^{\frac{3}{2}}\times 1=0$$

Multiply x and x to get x^{2}.

$$10\sqrt{13}x^{2}-5\sqrt{3}x+13^{\frac{3}{2}}\times 1=0$$

Reorder the terms.

$$10\sqrt{13}x^{2}-5\sqrt{3}x+13^{\frac{3}{2}}=0$$

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.

$$10\sqrt{13}x^{2}+\left(-5\sqrt{3}\right)x+13\sqrt{13}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10\sqrt{13} for a, -5\sqrt{3} for b, and 13\sqrt{13} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\sqrt{3}\right)±\sqrt{\left(-5\sqrt{3}\right)^{2}-4\times 10\sqrt{13}\times 13\sqrt{13}}}{2\times 10\sqrt{13}}$$

Square -5\sqrt{3}.

$$x=\frac{-\left(-5\sqrt{3}\right)±\sqrt{75-4\times 10\sqrt{13}\times 13\sqrt{13}}}{2\times 10\sqrt{13}}$$

Multiply -4 times 10\sqrt{13}.

$$x=\frac{-\left(-5\sqrt{3}\right)±\sqrt{75+\left(-40\sqrt{13}\right)\times 13\sqrt{13}}}{2\times 10\sqrt{13}}$$

Multiply -40\sqrt{13} times 13\sqrt{13}.

$$x=\frac{-\left(-5\sqrt{3}\right)±\sqrt{75-6760}}{2\times 10\sqrt{13}}$$

Add 75 to -6760.

$$x=\frac{-\left(-5\sqrt{3}\right)±\sqrt{-6685}}{2\times 10\sqrt{13}}$$

Take the square root of -6685.

$$x=\frac{-\left(-5\sqrt{3}\right)±\sqrt{6685}i}{2\times 10\sqrt{13}}$$

The opposite of -5\sqrt{3} is 5\sqrt{3}.

$$x=\frac{5\sqrt{3}±\sqrt{6685}i}{2\times 10\sqrt{13}}$$

Multiply 2 times 10\sqrt{13}.

$$x=\frac{5\sqrt{3}±\sqrt{6685}i}{20\sqrt{13}}$$

Now solve the equation x=\frac{5\sqrt{3}±\sqrt{6685}i}{20\sqrt{13}} when ± is plus. Add 5\sqrt{3} to i\sqrt{6685}.

$$x=\frac{5\sqrt{3}+\sqrt{6685}i}{20\sqrt{13}}$$

Divide 5\sqrt{3}+i\sqrt{6685} by 20\sqrt{13}.

$$x=\frac{\sqrt{13}\left(5\sqrt{3}+\sqrt{6685}i\right)}{260}$$

Now solve the equation x=\frac{5\sqrt{3}±\sqrt{6685}i}{20\sqrt{13}} when ± is minus. Subtract i\sqrt{6685} from 5\sqrt{3}.

$$x=\frac{-\sqrt{6685}i+5\sqrt{3}}{20\sqrt{13}}$$

Divide 5\sqrt{3}-i\sqrt{6685} by 20\sqrt{13}.

$$x=\frac{\sqrt{13}\left(-\sqrt{6685}i+5\sqrt{3}\right)}{260}$$

The equation is now solved.

$$x=\frac{\sqrt{13}\left(5\sqrt{3}+\sqrt{6685}i\right)}{260} x=\frac{\sqrt{13}\left(-\sqrt{6685}i+5\sqrt{3}\right)}{260}$$

Steps for Completing the Square

Multiply both sides of the equation by 26.

$$\frac{1}{400}x^{-1}\times 13^{\frac{1}{2}}\left(\left(20x\right)^{2}+5200+\left(60x\right)^{2}\right)=5\sqrt{3}$$

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

$$\frac{1}{400}x^{-1}\times 13^{\frac{1}{2}}\left(20^{2}x^{2}+5200+\left(60x\right)^{2}\right)=5\sqrt{3}$$

Calculate 20 to the power of 2 and get 400.

$$\frac{1}{400}x^{-1}\times 13^{\frac{1}{2}}\left(400x^{2}+5200+\left(60x\right)^{2}\right)=5\sqrt{3}$$

Expand \left(60x\right)^{2}.

$$\frac{1}{400}x^{-1}\times 13^{\frac{1}{2}}\left(400x^{2}+5200+60^{2}x^{2}\right)=5\sqrt{3}$$

Calculate 60 to the power of 2 and get 3600.

$$\frac{1}{400}x^{-1}\times 13^{\frac{1}{2}}\left(400x^{2}+5200+3600x^{2}\right)=5\sqrt{3}$$

Combine 400x^{2} and 3600x^{2} to get 4000x^{2}.

$$\frac{1}{400}x^{-1}\times 13^{\frac{1}{2}}\left(4000x^{2}+5200\right)=5\sqrt{3}$$

Use the distributive property to multiply \frac{1}{400}x^{-1}\times 13^{\frac{1}{2}} by 4000x^{2}+5200.

$$10x\times 13^{\frac{1}{2}}+13x^{-1}\times 13^{\frac{1}{2}}=5\sqrt{3}$$

To multiply powers of the same base, add their exponents. Add 1 and \frac{1}{2} to get \frac{3}{2}.

$$10x\times 13^{\frac{1}{2}}+13^{\frac{3}{2}}x^{-1}=5\sqrt{3}$$

Reorder the terms.

$$10\sqrt{13}x+13^{\frac{3}{2}}\times \frac{1}{x}=5\sqrt{3}$$

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$10\sqrt{13}xx+13^{\frac{3}{2}}\times 1=5\sqrt{3}x$$

Multiply x and x to get x^{2}.

$$10\sqrt{13}x^{2}+13^{\frac{3}{2}}\times 1=5\sqrt{3}x$$

Subtract 5\sqrt{3}x from both sides.

$$10\sqrt{13}x^{2}+13^{\frac{3}{2}}\times 1-5\sqrt{3}x=0$$

Reorder the terms.

$$10\sqrt{13}x^{2}+13^{\frac{3}{2}}-5\sqrt{3}x=0$$

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.

$$10\sqrt{13}x^{2}+\left(-5\sqrt{3}\right)x+13\sqrt{13}=0$$

Subtract 13^{\frac{3}{2}} from both sides of the equation.

$$10\sqrt{13}x^{2}+\left(-5\sqrt{3}\right)x+13\sqrt{13}-13\sqrt{13}=-13\sqrt{13}$$

Subtracting 13^{\frac{3}{2}} from itself leaves 0.

$$10\sqrt{13}x^{2}+\left(-5\sqrt{3}\right)x=-13\sqrt{13}$$

Divide both sides by 10\sqrt{13}.

$$\frac{10\sqrt{13}x^{2}+\left(-5\sqrt{3}\right)x}{10\sqrt{13}}=-\frac{13\sqrt{13}}{10\sqrt{13}}$$

Dividing by 10\sqrt{13} undoes the multiplication by 10\sqrt{13}.

$$x^{2}+\left(-\frac{5\sqrt{3}}{10\sqrt{13}}\right)x=-\frac{13\sqrt{13}}{10\sqrt{13}}$$

Divide -5\sqrt{3} by 10\sqrt{13}.

$$x^{2}+\left(-\frac{\sqrt{39}}{26}\right)x=-\frac{13\sqrt{13}}{10\sqrt{13}}$$

Divide -13\sqrt{13} by 10\sqrt{13}.

$$x^{2}+\left(-\frac{\sqrt{39}}{26}\right)x=-\frac{13}{10}$$

Divide -\frac{\sqrt{39}}{26}, the coefficient of the x term, by 2 to get -\frac{\sqrt{39}}{52}. Then add the square of -\frac{\sqrt{39}}{52} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\left(-\frac{\sqrt{39}}{26}\right)x+\left(-\frac{\sqrt{39}}{52}\right)^{2}=-\frac{13}{10}+\left(-\frac{\sqrt{39}}{52}\right)^{2}$$

Square -\frac{\sqrt{39}}{52}.

$$x^{2}+\left(-\frac{\sqrt{39}}{26}\right)x+\frac{3}{208}=-\frac{13}{10}+\frac{3}{208}$$

Add -\frac{13}{10} to \frac{3}{208} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\left(-\frac{\sqrt{39}}{26}\right)x+\frac{3}{208}=-\frac{1337}{1040}$$

Factor x^{2}+\left(-\frac{\sqrt{39}}{26}\right)x+\frac{3}{208}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{\sqrt{39}}{52}\right)^{2}=-\frac{1337}{1040}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{\sqrt{39}}{52}\right)^{2}}=\sqrt{-\frac{1337}{1040}}$$

Simplify.

$$x-\frac{\sqrt{39}}{52}=\frac{\sqrt{86905}i}{260} x-\frac{\sqrt{39}}{52}=-\frac{\sqrt{86905}i}{260}$$

Add \frac{\sqrt{39}}{52} to both sides of the equation.

$$x=\frac{\sqrt{86905}i}{260}+\frac{\sqrt{39}}{52} x=-\frac{\sqrt{86905}i}{260}+\frac{\sqrt{39}}{52}$$