Solve 5x-9x^2=3x-4x^2

$$5 x - 9 x ^ { 2 } = 3 x - 4 x ^ { 2 }$$

Solve for x

$$x=\frac{2}{5}=0.4$$

$$x=0$$

Steps Using Factoring

Subtract 3x from both sides.

$$5x-9x^{2}-3x=-4x^{2}$$

Combine 5x and -3x to get 2x.

$$2x-9x^{2}=-4x^{2}$$

Add 4x^{2} to both sides.

$$2x-9x^{2}+4x^{2}=0$$

Combine -9x^{2} and 4x^{2} to get -5x^{2}.

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

Factor out x.

$$x\left(2-5x\right)=0$$

To find equation solutions, solve x=0 and 2-5x=0.

$$x=0 x=\frac{2}{5}$$

Steps Using the Quadratic Formula

Subtract 3x from both sides.

$$5x-9x^{2}-3x=-4x^{2}$$

Combine 5x and -3x to get 2x.

$$2x-9x^{2}=-4x^{2}$$

Add 4x^{2} to both sides.

$$2x-9x^{2}+4x^{2}=0$$

Combine -9x^{2} and 4x^{2} to get -5x^{2}.

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

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.

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

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

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

Take the square root of 2^{2}.

$$x=\frac{-2±2}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-2±2}{-10}$$

Now solve the equation x=\frac{-2±2}{-10} when ± is plus. Add -2 to 2.

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

Divide 0 by -10.

$$x=0$$

Now solve the equation x=\frac{-2±2}{-10} when ± is minus. Subtract 2 from -2.

$$x=-\frac{4}{-10}$$

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

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

The equation is now solved.

$$x=0 x=\frac{2}{5}$$

Steps for Completing the Square

Subtract 3x from both sides.

$$5x-9x^{2}-3x=-4x^{2}$$

Combine 5x and -3x to get 2x.

$$2x-9x^{2}=-4x^{2}$$

Add 4x^{2} to both sides.

$$2x-9x^{2}+4x^{2}=0$$

Combine -9x^{2} and 4x^{2} to get -5x^{2}.

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

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.

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

Divide both sides by -5.

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

Dividing by -5 undoes the multiplication by -5.

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

Divide 2 by -5.

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

Divide 0 by -5.

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

Divide -\frac{2}{5}, the coefficient of the x term, by 2 to get -\frac{1}{5}. Then add the square of -\frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{5}x+\left(-\frac{1}{5}\right)^{2}=\left(-\frac{1}{5}\right)^{2}$$

Square -\frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{1}{25}$$

Factor x^{2}-\frac{2}{5}x+\frac{1}{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}.

$$\left(x-\frac{1}{5}\right)^{2}=\frac{1}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{\frac{1}{25}}$$

Simplify.

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

Add \frac{1}{5} to both sides of the equation.

$$x=\frac{2}{5} x=0$$