Solve -5x=3x+4/2x-5

$$- 5 x = \frac { 3 x + 4 } { 2 x - 5 }$$

Solve for x

$$x=2$$

$$x=\frac{1}{5}=0.2$$

Steps Using the Quadratic Formula

Variable x cannot be equal to \frac{5}{2} since division by zero is not defined. Multiply both sides of the equation by 2x-5.

$$-5x\left(2x-5\right)=3x+4$$

Use the distributive property to multiply -5x by 2x-5.

$$-10x^{2}+25x=3x+4$$

Subtract 3x from both sides.

$$-10x^{2}+25x-3x=4$$

Combine 25x and -3x to get 22x.

$$-10x^{2}+22x=4$$

Subtract 4 from both sides.

$$-10x^{2}+22x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 22 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-22±\sqrt{22^{2}-4\left(-10\right)\left(-4\right)}}{2\left(-10\right)}$$

Square 22.

$$x=\frac{-22±\sqrt{484-4\left(-10\right)\left(-4\right)}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-22±\sqrt{484+40\left(-4\right)}}{2\left(-10\right)}$$

Multiply 40 times -4.

$$x=\frac{-22±\sqrt{484-160}}{2\left(-10\right)}$$

Add 484 to -160.

$$x=\frac{-22±\sqrt{324}}{2\left(-10\right)}$$

Take the square root of 324.

$$x=\frac{-22±18}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{-22±18}{-20}$$

Now solve the equation x=\frac{-22±18}{-20} when ± is plus. Add -22 to 18.

$$x=-\frac{4}{-20}$$

Reduce the fraction \frac{-4}{-20} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{5}$$

Now solve the equation x=\frac{-22±18}{-20} when ± is minus. Subtract 18 from -22.

$$x=-\frac{40}{-20}$$

Divide -40 by -20.

$$x=2$$

The equation is now solved.

$$x=\frac{1}{5} x=2$$

Steps for Completing the Square

Variable x cannot be equal to \frac{5}{2} since division by zero is not defined. Multiply both sides of the equation by 2x-5.

$$-5x\left(2x-5\right)=3x+4$$

Use the distributive property to multiply -5x by 2x-5.

$$-10x^{2}+25x=3x+4$$

Subtract 3x from both sides.

$$-10x^{2}+25x-3x=4$$

Combine 25x and -3x to get 22x.

$$-10x^{2}+22x=4$$

Divide both sides by -10.

$$\frac{-10x^{2}+22x}{-10}=\frac{4}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$x^{2}+\frac{22}{-10}x=\frac{4}{-10}$$

Reduce the fraction \frac{22}{-10} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{11}{5}x=\frac{4}{-10}$$

Reduce the fraction \frac{4}{-10} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{11}{5}x=-\frac{2}{5}$$

Divide -\frac{11}{5}, the coefficient of the x term, by 2 to get -\frac{11}{10}. Then add the square of -\frac{11}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{5}x+\left(-\frac{11}{10}\right)^{2}=-\frac{2}{5}+\left(-\frac{11}{10}\right)^{2}$$

Square -\frac{11}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{5}x+\frac{121}{100}=-\frac{2}{5}+\frac{121}{100}$$

Add -\frac{2}{5} to \frac{121}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{5}x+\frac{121}{100}=\frac{81}{100}$$

Factor x^{2}-\frac{11}{5}x+\frac{121}{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}.

$$\left(x-\frac{11}{10}\right)^{2}=\frac{81}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{10}\right)^{2}}=\sqrt{\frac{81}{100}}$$

Simplify.

$$x-\frac{11}{10}=\frac{9}{10} x-\frac{11}{10}=-\frac{9}{10}$$

Add \frac{11}{10} to both sides of the equation.

$$x=2 x=\frac{1}{5}$$