Solve (4x+3)(2-3x)=6

$$(4x+3)(2-3x)=6$$

Solve for x

$$x=-\frac{1}{12}\approx -0.083333333$$

$$x=0$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4x+3 by 2-3x and combine like terms.

$$-x-12x^{2}+6=6$$

Subtract 6 from both sides.

$$-x-12x^{2}+6-6=0$$

Subtract 6 from 6 to get 0.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -12 for a, -1 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1}}{2\left(-12\right)}$$

Take the square root of 1.

$$x=\frac{-\left(-1\right)±1}{2\left(-12\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±1}{2\left(-12\right)}$$

Multiply 2 times -12.

$$x=\frac{1±1}{-24}$$

Now solve the equation x=\frac{1±1}{-24} when ± is plus. Add 1 to 1.

$$x=\frac{2}{-24}$$

Reduce the fraction \frac{2}{-24} to lowest terms by extracting and canceling out 2.

$$x=-\frac{1}{12}$$

Now solve the equation x=\frac{1±1}{-24} when ± is minus. Subtract 1 from 1.

$$x=\frac{0}{-24}$$

Divide 0 by -24.

$$x=0$$

The equation is now solved.

$$x=-\frac{1}{12} x=0$$

Steps for Completing the Square

Use the distributive property to multiply 4x+3 by 2-3x and combine like terms.

$$-x-12x^{2}+6=6$$

Subtract 6 from both sides.

$$-x-12x^{2}=6-6$$

Subtract 6 from 6 to get 0.

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

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.

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

Divide both sides by -12.

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

Dividing by -12 undoes the multiplication by -12.

$$x^{2}+\left(-\frac{1}{-12}\right)x=\frac{0}{-12}$$

Divide -1 by -12.

$$x^{2}+\frac{1}{12}x=\frac{0}{-12}$$

Divide 0 by -12.

$$x^{2}+\frac{1}{12}x=0$$

Divide \frac{1}{12}, the coefficient of the x term, by 2 to get \frac{1}{24}. Then add the square of \frac{1}{24} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{12}x+\left(\frac{1}{24}\right)^{2}=\left(\frac{1}{24}\right)^{2}$$

Square \frac{1}{24} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{12}x+\frac{1}{576}=\frac{1}{576}$$

Factor x^{2}+\frac{1}{12}x+\frac{1}{576}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{24}\right)^{2}=\frac{1}{576}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{24}\right)^{2}}=\sqrt{\frac{1}{576}}$$

Simplify.

$$x+\frac{1}{24}=\frac{1}{24} x+\frac{1}{24}=-\frac{1}{24}$$

Subtract \frac{1}{24} from both sides of the equation.

$$x=0 x=-\frac{1}{12}$$