Solve (-6x+5)(2x-3)=-15

$$( - 6 x + 5 ) ( 2 x - 3 ) = - 15$$

Solve for x

$$x = \frac{7}{3} = 2\frac{1}{3} \approx 2.333333333$$

$$x=0$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -6x+5 by 2x-3 and combine like terms.

$$-12x^{2}+28x-15=-15$$

Add 15 to both sides.

$$-12x^{2}+28x-15+15=0$$

Add -15 and 15 to get 0.

$$-12x^{2}+28x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -12 for a, 28 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-28±\sqrt{28^{2}}}{2\left(-12\right)}$$

Take the square root of 28^{2}.

$$x=\frac{-28±28}{2\left(-12\right)}$$

Multiply 2 times -12.

$$x=\frac{-28±28}{-24}$$

Now solve the equation x=\frac{-28±28}{-24} when ± is plus. Add -28 to 28.

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

Divide 0 by -24.

$$x=0$$

Now solve the equation x=\frac{-28±28}{-24} when ± is minus. Subtract 28 from -28.

$$x=-\frac{56}{-24}$$

Reduce the fraction \frac{-56}{-24} to lowest terms by extracting and canceling out 8.

$$x=\frac{7}{3}$$

The equation is now solved.

$$x=0 x=\frac{7}{3}$$

Steps for Completing the Square

Use the distributive property to multiply -6x+5 by 2x-3 and combine like terms.

$$-12x^{2}+28x-15=-15$$

Add 15 to both sides.

$$-12x^{2}+28x=-15+15$$

Add -15 and 15 to get 0.

$$-12x^{2}+28x=0$$

Divide both sides by -12.

$$\frac{-12x^{2}+28x}{-12}=\frac{0}{-12}$$

Dividing by -12 undoes the multiplication by -12.

$$x^{2}+\frac{28}{-12}x=\frac{0}{-12}$$

Reduce the fraction \frac{28}{-12} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{7}{3}x=\frac{0}{-12}$$

Divide 0 by -12.

$$x^{2}-\frac{7}{3}x=0$$

Divide -\frac{7}{3}, the coefficient of the x term, by 2 to get -\frac{7}{6}. Then add the square of -\frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{3}x+\left(-\frac{7}{6}\right)^{2}=\left(-\frac{7}{6}\right)^{2}$$

Square -\frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{3}x+\frac{49}{36}=\frac{49}{36}$$

Factor x^{2}-\frac{7}{3}x+\frac{49}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{6}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$x-\frac{7}{6}=\frac{7}{6} x-\frac{7}{6}=-\frac{7}{6}$$

Add \frac{7}{6} to both sides of the equation.

$$x=\frac{7}{3} x=0$$