Solve 5x+5x*(x-6)=3x

$$5 x + 5 x \cdot ( x - 6 ) = 3 x$$

Solve for x

$$x = \frac{28}{5} = 5\frac{3}{5} = 5.6$$

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply 5x by x-6.

$$5x+5x^{2}-30x=3x$$

Combine 5x and -30x to get -25x.

$$-25x+5x^{2}=3x$$

Subtract 3x from both sides.

$$-25x+5x^{2}-3x=0$$

Combine -25x and -3x to get -28x.

$$-28x+5x^{2}=0$$

Factor out x.

$$x\left(-28+5x\right)=0$$

To find equation solutions, solve x=0 and -28+5x=0.

$$x=0 x=\frac{28}{5}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 5x by x-6.

$$5x+5x^{2}-30x=3x$$

Combine 5x and -30x to get -25x.

$$-25x+5x^{2}=3x$$

Subtract 3x from both sides.

$$-25x+5x^{2}-3x=0$$

Combine -25x and -3x to get -28x.

$$-28x+5x^{2}=0$$

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.

$$5x^{2}-28x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -28 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}}}{2\times 5}$$

Take the square root of \left(-28\right)^{2}.

$$x=\frac{-\left(-28\right)±28}{2\times 5}$$

The opposite of -28 is 28.

$$x=\frac{28±28}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{28±28}{10}$$

Now solve the equation x=\frac{28±28}{10} when ± is plus. Add 28 to 28.

$$x=\frac{56}{10}$$

Reduce the fraction \frac{56}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{28}{5}$$

Now solve the equation x=\frac{28±28}{10} when ± is minus. Subtract 28 from 28.

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

Divide 0 by 10.

$$x=0$$

The equation is now solved.

$$x=\frac{28}{5} x=0$$

Steps for Completing the Square

Use the distributive property to multiply 5x by x-6.

$$5x+5x^{2}-30x=3x$$

Combine 5x and -30x to get -25x.

$$-25x+5x^{2}=3x$$

Subtract 3x from both sides.

$$-25x+5x^{2}-3x=0$$

Combine -25x and -3x to get -28x.

$$-28x+5x^{2}=0$$

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.

$$5x^{2}-28x=0$$

Divide both sides by 5.

$$\frac{5x^{2}-28x}{5}=\frac{0}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{28}{5}x=\frac{0}{5}$$

Divide 0 by 5.

$$x^{2}-\frac{28}{5}x=0$$

Divide -\frac{28}{5}, the coefficient of the x term, by 2 to get -\frac{14}{5}. Then add the square of -\frac{14}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{28}{5}x+\left(-\frac{14}{5}\right)^{2}=\left(-\frac{14}{5}\right)^{2}$$

Square -\frac{14}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{28}{5}x+\frac{196}{25}=\frac{196}{25}$$

Factor x^{2}-\frac{28}{5}x+\frac{196}{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}.

$$\left(x-\frac{14}{5}\right)^{2}=\frac{196}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{14}{5}\right)^{2}}=\sqrt{\frac{196}{25}}$$

Simplify.

$$x-\frac{14}{5}=\frac{14}{5} x-\frac{14}{5}=-\frac{14}{5}$$

Add \frac{14}{5} to both sides of the equation.

$$x=\frac{28}{5} x=0$$