Solve -3x(2-x+4x)=1

$$- 3 x ( 2 - x + 4 x ) = 1$$

Solve for x

$$x=-\frac{1}{3}\approx -0.333333333$$

Steps Using the Quadratic Formula

Combine -x and 4x to get 3x.

$$-3x\left(2+3x\right)=1$$

Use the distributive property to multiply -3x by 2+3x.

$$-6x-9x^{2}=1$$

Subtract 1 from both sides.

$$-6x-9x^{2}-1=0$$

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.

$$-9x^{2}-6x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, -6 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-9\right)\left(-1\right)}}{2\left(-9\right)}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-9\right)\left(-1\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-\left(-6\right)±\sqrt{36+36\left(-1\right)}}{2\left(-9\right)}$$

Multiply 36 times -1.

$$x=\frac{-\left(-6\right)±\sqrt{36-36}}{2\left(-9\right)}$$

Add 36 to -36.

$$x=\frac{-\left(-6\right)±\sqrt{0}}{2\left(-9\right)}$$

Take the square root of 0.

$$x=-\frac{-6}{2\left(-9\right)}$$

The opposite of -6 is 6.

$$x=\frac{6}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{6}{-18}$$

Reduce the fraction \frac{6}{-18} to lowest terms by extracting and canceling out 6.

$$x=-\frac{1}{3}$$

Steps for Completing the Square

Combine -x and 4x to get 3x.

$$-3x\left(2+3x\right)=1$$

Use the distributive property to multiply -3x by 2+3x.

$$-6x-9x^{2}=1$$

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.

$$-9x^{2}-6x=1$$

Divide both sides by -9.

$$\frac{-9x^{2}-6x}{-9}=\frac{1}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$x^{2}+\left(-\frac{6}{-9}\right)x=\frac{1}{-9}$$

Reduce the fraction \frac{-6}{-9} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{2}{3}x=\frac{1}{-9}$$

Divide 1 by -9.

$$x^{2}+\frac{2}{3}x=-\frac{1}{9}$$

Divide \frac{2}{3}, the coefficient of the x term, by 2 to get \frac{1}{3}. Then add the square of \frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=-\frac{1}{9}+\left(\frac{1}{3}\right)^{2}$$

Square \frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{-1+1}{9}$$

Add -\frac{1}{9} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=0$$

Factor x^{2}+\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{3}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+\frac{1}{3}=0 x+\frac{1}{3}=0$$

Subtract \frac{1}{3} from both sides of the equation.

$$x=-\frac{1}{3} x=-\frac{1}{3}$$

The equation is now solved. Solutions are the same.

$$x=-\frac{1}{3}$$