Solve 4x^2-5x=7x-x^2

$$4 x ^ { 2 } - 5 x = 7 x - x ^ { 2 }$$

Solve for x

$$x = \frac{12}{5} = 2\frac{2}{5} = 2.4$$

$$x=0$$

Steps Using Factoring

Subtract 7x from both sides.

$$4x^{2}-5x-7x=-x^{2}$$

Combine -5x and -7x to get -12x.

$$4x^{2}-12x=-x^{2}$$

Add x^{2} to both sides.

$$4x^{2}-12x+x^{2}=0$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$5x^{2}-12x=0$$

Factor out x.

$$x\left(5x-12\right)=0$$

To find equation solutions, solve x=0 and 5x-12=0.

$$x=0 x=\frac{12}{5}$$

Steps Using the Quadratic Formula

Subtract 7x from both sides.

$$4x^{2}-5x-7x=-x^{2}$$

Combine -5x and -7x to get -12x.

$$4x^{2}-12x=-x^{2}$$

Add x^{2} to both sides.

$$4x^{2}-12x+x^{2}=0$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$5x^{2}-12x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -12 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}}}{2\times 5}$$

Take the square root of \left(-12\right)^{2}.

$$x=\frac{-\left(-12\right)±12}{2\times 5}$$

The opposite of -12 is 12.

$$x=\frac{12±12}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{12±12}{10}$$

Now solve the equation x=\frac{12±12}{10} when ± is plus. Add 12 to 12.

$$x=\frac{24}{10}$$

Reduce the fraction \frac{24}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{12}{5}$$

Now solve the equation x=\frac{12±12}{10} when ± is minus. Subtract 12 from 12.

$$x=\frac{0}{10}$$

Divide 0 by 10.

$$x=0$$

The equation is now solved.

$$x=\frac{12}{5} x=0$$

Steps for Completing the Square

Subtract 7x from both sides.

$$4x^{2}-5x-7x=-x^{2}$$

Combine -5x and -7x to get -12x.

$$4x^{2}-12x=-x^{2}$$

Add x^{2} to both sides.

$$4x^{2}-12x+x^{2}=0$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$5x^{2}-12x=0$$

Divide both sides by 5.

$$\frac{5x^{2}-12x}{5}=\frac{0}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{12}{5}x=\frac{0}{5}$$

Divide 0 by 5.

$$x^{2}-\frac{12}{5}x=0$$

Divide -\frac{12}{5}, the coefficient of the x term, by 2 to get -\frac{6}{5}. Then add the square of -\frac{6}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{12}{5}x+\left(-\frac{6}{5}\right)^{2}=\left(-\frac{6}{5}\right)^{2}$$

Square -\frac{6}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{12}{5}x+\frac{36}{25}=\frac{36}{25}$$

Factor x^{2}-\frac{12}{5}x+\frac{36}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{6}{5}\right)^{2}=\frac{36}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{6}{5}\right)^{2}}=\sqrt{\frac{36}{25}}$$

Simplify.

$$x-\frac{6}{5}=\frac{6}{5} x-\frac{6}{5}=-\frac{6}{5}$$

Add \frac{6}{5} to both sides of the equation.

$$x=\frac{12}{5} x=0$$