Solve 594/x-648/x-3=5

$$\frac{ 594 }{ x } - \frac{ 648 }{ x-3 } =5$$

Solve for x (complex solution)

$$x=\frac{-3\sqrt{3791}i-39}{10}\approx -3.9-18.471329135i$$

$$x=\frac{-39+3\sqrt{3791}i}{10}\approx -3.9+18.471329135i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by x\left(x-3\right), the least common multiple of x,x-3.

$$\left(x-3\right)\times 594-x\times 648=5x\left(x-3\right)$$

Use the distributive property to multiply x-3 by 594.

$$594x-1782-x\times 648=5x\left(x-3\right)$$

Use the distributive property to multiply 5x by x-3.

$$594x-1782-x\times 648=5x^{2}-15x$$

Subtract 5x^{2} from both sides.

$$594x-1782-x\times 648-5x^{2}=-15x$$

Add 15x to both sides.

$$594x-1782-x\times 648-5x^{2}+15x=0$$

Combine 594x and 15x to get 609x.

$$609x-1782-x\times 648-5x^{2}=0$$

Multiply -1 and 648 to get -648.

$$609x-1782-648x-5x^{2}=0$$

Combine 609x and -648x to get -39x.

$$-39x-1782-5x^{2}=0$$

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.

$$-5x^{2}-39x-1782=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -39 for b, and -1782 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-39\right)±\sqrt{\left(-39\right)^{2}-4\left(-5\right)\left(-1782\right)}}{2\left(-5\right)}$$

Square -39.

$$x=\frac{-\left(-39\right)±\sqrt{1521-4\left(-5\right)\left(-1782\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-\left(-39\right)±\sqrt{1521+20\left(-1782\right)}}{2\left(-5\right)}$$

Multiply 20 times -1782.

$$x=\frac{-\left(-39\right)±\sqrt{1521-35640}}{2\left(-5\right)}$$

Add 1521 to -35640.

$$x=\frac{-\left(-39\right)±\sqrt{-34119}}{2\left(-5\right)}$$

Take the square root of -34119.

$$x=\frac{-\left(-39\right)±3\sqrt{3791}i}{2\left(-5\right)}$$

The opposite of -39 is 39.

$$x=\frac{39±3\sqrt{3791}i}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{39±3\sqrt{3791}i}{-10}$$

Now solve the equation x=\frac{39±3\sqrt{3791}i}{-10} when ± is plus. Add 39 to 3i\sqrt{3791}.

$$x=\frac{39+3\sqrt{3791}i}{-10}$$

Divide 39+3i\sqrt{3791} by -10.

$$x=\frac{-3\sqrt{3791}i-39}{10}$$

Now solve the equation x=\frac{39±3\sqrt{3791}i}{-10} when ± is minus. Subtract 3i\sqrt{3791} from 39.

$$x=\frac{-3\sqrt{3791}i+39}{-10}$$

Divide 39-3i\sqrt{3791} by -10.

$$x=\frac{-39+3\sqrt{3791}i}{10}$$

The equation is now solved.

$$x=\frac{-3\sqrt{3791}i-39}{10} x=\frac{-39+3\sqrt{3791}i}{10}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by x\left(x-3\right), the least common multiple of x,x-3.

$$\left(x-3\right)\times 594-x\times 648=5x\left(x-3\right)$$

Use the distributive property to multiply x-3 by 594.

$$594x-1782-x\times 648=5x\left(x-3\right)$$

Use the distributive property to multiply 5x by x-3.

$$594x-1782-x\times 648=5x^{2}-15x$$

Subtract 5x^{2} from both sides.

$$594x-1782-x\times 648-5x^{2}=-15x$$

Add 15x to both sides.

$$594x-1782-x\times 648-5x^{2}+15x=0$$

Combine 594x and 15x to get 609x.

$$609x-1782-x\times 648-5x^{2}=0$$

Add 1782 to both sides. Anything plus zero gives itself.

$$609x-x\times 648-5x^{2}=1782$$

Multiply -1 and 648 to get -648.

$$609x-648x-5x^{2}=1782$$

Combine 609x and -648x to get -39x.

$$-39x-5x^{2}=1782$$

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.

$$-5x^{2}-39x=1782$$

Divide both sides by -5.

$$\frac{-5x^{2}-39x}{-5}=\frac{1782}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\left(-\frac{39}{-5}\right)x=\frac{1782}{-5}$$

Divide -39 by -5.

$$x^{2}+\frac{39}{5}x=\frac{1782}{-5}$$

Divide 1782 by -5.

$$x^{2}+\frac{39}{5}x=-\frac{1782}{5}$$

Divide \frac{39}{5}, the coefficient of the x term, by 2 to get \frac{39}{10}. Then add the square of \frac{39}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{39}{5}x+\left(\frac{39}{10}\right)^{2}=-\frac{1782}{5}+\left(\frac{39}{10}\right)^{2}$$

Square \frac{39}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{39}{5}x+\frac{1521}{100}=-\frac{1782}{5}+\frac{1521}{100}$$

Add -\frac{1782}{5} to \frac{1521}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{39}{5}x+\frac{1521}{100}=-\frac{34119}{100}$$

Factor x^{2}+\frac{39}{5}x+\frac{1521}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{39}{10}\right)^{2}=-\frac{34119}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{39}{10}\right)^{2}}=\sqrt{-\frac{34119}{100}}$$

Simplify.

$$x+\frac{39}{10}=\frac{3\sqrt{3791}i}{10} x+\frac{39}{10}=-\frac{3\sqrt{3791}i}{10}$$

Subtract \frac{39}{10} from both sides of the equation.

$$x=\frac{-39+3\sqrt{3791}i}{10} x=\frac{-3\sqrt{3791}i-39}{10}$$