Solve 4x^2-5x=6x

$$4 x ^ { 2 } - 5 x = 6 x$$

Solve for x

$$x = \frac{11}{4} = 2\frac{3}{4} = 2.75$$

$$x=0$$

Steps Using Factoring

Subtract 6x from both sides.

$$4x^{2}-5x-6x=0$$

Combine -5x and -6x to get -11x.

$$4x^{2}-11x=0$$

Factor out x.

$$x\left(4x-11\right)=0$$

To find equation solutions, solve x=0 and 4x-11=0.

$$x=0 x=\frac{11}{4}$$

Steps Using the Quadratic Formula

Subtract 6x from both sides.

$$4x^{2}-5x-6x=0$$

Combine -5x and -6x to get -11x.

$$4x^{2}-11x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -11 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}}}{2\times 4}$$

Take the square root of \left(-11\right)^{2}.

$$x=\frac{-\left(-11\right)±11}{2\times 4}$$

The opposite of -11 is 11.

$$x=\frac{11±11}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{11±11}{8}$$

Now solve the equation x=\frac{11±11}{8} when ± is plus. Add 11 to 11.

$$x=\frac{22}{8}$$

Reduce the fraction \frac{22}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{11}{4}$$

Now solve the equation x=\frac{11±11}{8} when ± is minus. Subtract 11 from 11.

$$x=\frac{0}{8}$$

Divide 0 by 8.

$$x=0$$

The equation is now solved.

$$x=\frac{11}{4} x=0$$

Steps for Completing the Square

Subtract 6x from both sides.

$$4x^{2}-5x-6x=0$$

Combine -5x and -6x to get -11x.

$$4x^{2}-11x=0$$

Divide both sides by 4.

$$\frac{4x^{2}-11x}{4}=\frac{0}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}-\frac{11}{4}x=\frac{0}{4}$$

Divide 0 by 4.

$$x^{2}-\frac{11}{4}x=0$$

Divide -\frac{11}{4}, the coefficient of the x term, by 2 to get -\frac{11}{8}. Then add the square of -\frac{11}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{4}x+\left(-\frac{11}{8}\right)^{2}=\left(-\frac{11}{8}\right)^{2}$$

Square -\frac{11}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{4}x+\frac{121}{64}=\frac{121}{64}$$

Factor x^{2}-\frac{11}{4}x+\frac{121}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{8}\right)^{2}=\frac{121}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{8}\right)^{2}}=\sqrt{\frac{121}{64}}$$

Simplify.

$$x-\frac{11}{8}=\frac{11}{8} x-\frac{11}{8}=-\frac{11}{8}$$

Add \frac{11}{8} to both sides of the equation.

$$x=\frac{11}{4} x=0$$