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

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

Solve for x

$$x=4$$

$$x=0$$

Steps Using Factoring

Subtract 7x from both sides.

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

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

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

Subtract x^{2} from both sides.

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

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-12x=0$$

Factor out x.

$$x\left(3x-12\right)=0$$

To find equation solutions, solve x=0 and 3x-12=0.

$$x=0 x=4$$

Steps Using the Quadratic Formula

Subtract 7x from both sides.

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

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

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

Subtract x^{2} from both sides.

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

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-12x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -12 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}}}{2\times 3}$$

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

$$x=\frac{-\left(-12\right)±12}{2\times 3}$$

The opposite of -12 is 12.

$$x=\frac{12±12}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{12±12}{6}$$

Now solve the equation x=\frac{12±12}{6} when ± is plus. Add 12 to 12.

$$x=\frac{24}{6}$$

Divide 24 by 6.

$$x=4$$

Now solve the equation x=\frac{12±12}{6} when ± is minus. Subtract 12 from 12.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

The equation is now solved.

$$x=4 x=0$$

Steps for Completing the Square

Subtract 7x from both sides.

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

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

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

Subtract x^{2} from both sides.

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

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-12x=0$$

Divide both sides by 3.

$$\frac{3x^{2}-12x}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{12}{3}\right)x=\frac{0}{3}$$

Divide -12 by 3.

$$x^{2}-4x=\frac{0}{3}$$

Divide 0 by 3.

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

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

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

Square -2.

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

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

$$\left(x-2\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-2=2 x-2=-2$$

Add 2 to both sides of the equation.

$$x=4 x=0$$