Solve -10x*(2x)-10x*(x)=3x

$$- 10 x \cdot ( 2 x ) - 10 x \cdot ( x ) = 3 x$$

Solve for x

$$x=-\frac{1}{10}=-0.1$$

$$x=0$$

Steps Using Factoring

Multiply x and x to get x^{2}.

$$-10x^{2}\times 2-10xx=3x$$

Multiply x and x to get x^{2}.

$$-10x^{2}\times 2-10x^{2}=3x$$

Multiply -10 and 2 to get -20.

$$-20x^{2}-10x^{2}=3x$$

Combine -20x^{2} and -10x^{2} to get -30x^{2}.

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

Subtract 3x from both sides.

$$-30x^{2}-3x=0$$

Factor out x.

$$x\left(-30x-3\right)=0$$

To find equation solutions, solve x=0 and -30x-3=0.

$$x=0 x=-\frac{1}{10}$$

Steps Using the Quadratic Formula

Multiply x and x to get x^{2}.

$$-10x^{2}\times 2-10xx=3x$$

Multiply x and x to get x^{2}.

$$-10x^{2}\times 2-10x^{2}=3x$$

Multiply -10 and 2 to get -20.

$$-20x^{2}-10x^{2}=3x$$

Combine -20x^{2} and -10x^{2} to get -30x^{2}.

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

Subtract 3x from both sides.

$$-30x^{2}-3x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -30 for a, -3 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}}}{2\left(-30\right)}$$

Take the square root of \left(-3\right)^{2}.

$$x=\frac{-\left(-3\right)±3}{2\left(-30\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±3}{2\left(-30\right)}$$

Multiply 2 times -30.

$$x=\frac{3±3}{-60}$$

Now solve the equation x=\frac{3±3}{-60} when ± is plus. Add 3 to 3.

$$x=\frac{6}{-60}$$

Reduce the fraction \frac{6}{-60} to lowest terms by extracting and canceling out 6.

$$x=-\frac{1}{10}$$

Now solve the equation x=\frac{3±3}{-60} when ± is minus. Subtract 3 from 3.

$$x=\frac{0}{-60}$$

Divide 0 by -60.

$$x=0$$

The equation is now solved.

$$x=-\frac{1}{10} x=0$$

Steps for Completing the Square

Multiply x and x to get x^{2}.

$$-10x^{2}\times 2-10xx=3x$$

Multiply x and x to get x^{2}.

$$-10x^{2}\times 2-10x^{2}=3x$$

Multiply -10 and 2 to get -20.

$$-20x^{2}-10x^{2}=3x$$

Combine -20x^{2} and -10x^{2} to get -30x^{2}.

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

Subtract 3x from both sides.

$$-30x^{2}-3x=0$$

Divide both sides by -30.

$$\frac{-30x^{2}-3x}{-30}=\frac{0}{-30}$$

Dividing by -30 undoes the multiplication by -30.

$$x^{2}+\left(-\frac{3}{-30}\right)x=\frac{0}{-30}$$

Reduce the fraction \frac{-3}{-30} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{1}{10}x=\frac{0}{-30}$$

Divide 0 by -30.

$$x^{2}+\frac{1}{10}x=0$$

Divide \frac{1}{10}, the coefficient of the x term, by 2 to get \frac{1}{20}. Then add the square of \frac{1}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{10}x+\left(\frac{1}{20}\right)^{2}=\left(\frac{1}{20}\right)^{2}$$

Square \frac{1}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{10}x+\frac{1}{400}=\frac{1}{400}$$

Factor x^{2}+\frac{1}{10}x+\frac{1}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{20}\right)^{2}=\frac{1}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{20}\right)^{2}}=\sqrt{\frac{1}{400}}$$

Simplify.

$$x+\frac{1}{20}=\frac{1}{20} x+\frac{1}{20}=-\frac{1}{20}$$

Subtract \frac{1}{20} from both sides of the equation.

$$x=0 x=-\frac{1}{10}$$