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

$$3 x ( x - 5 ) = 6 ( x - 5 )$$

Solve for x

$$x=2$$

$$x=5$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x by x-5.

$$3x^{2}-15x=6\left(x-5\right)$$

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

$$3x^{2}-15x=6x-30$$

Subtract 6x from both sides.

$$3x^{2}-15x-6x=-30$$

Combine -15x and -6x to get -21x.

$$3x^{2}-21x=-30$$

Add 30 to both sides.

$$3x^{2}-21x+30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -21 for b, and 30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 3\times 30}}{2\times 3}$$

Square -21.

$$x=\frac{-\left(-21\right)±\sqrt{441-4\times 3\times 30}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-21\right)±\sqrt{441-12\times 30}}{2\times 3}$$

Multiply -12 times 30.

$$x=\frac{-\left(-21\right)±\sqrt{441-360}}{2\times 3}$$

Add 441 to -360.

$$x=\frac{-\left(-21\right)±\sqrt{81}}{2\times 3}$$

Take the square root of 81.

$$x=\frac{-\left(-21\right)±9}{2\times 3}$$

The opposite of -21 is 21.

$$x=\frac{21±9}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{21±9}{6}$$

Now solve the equation x=\frac{21±9}{6} when ± is plus. Add 21 to 9.

$$x=\frac{30}{6}$$

Divide 30 by 6.

$$x=5$$

Now solve the equation x=\frac{21±9}{6} when ± is minus. Subtract 9 from 21.

$$x=\frac{12}{6}$$

Divide 12 by 6.

$$x=2$$

The equation is now solved.

$$x=5 x=2$$

Steps for Completing the Square

Use the distributive property to multiply 3x by x-5.

$$3x^{2}-15x=6\left(x-5\right)$$

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

$$3x^{2}-15x=6x-30$$

Subtract 6x from both sides.

$$3x^{2}-15x-6x=-30$$

Combine -15x and -6x to get -21x.

$$3x^{2}-21x=-30$$

Divide both sides by 3.

$$\frac{3x^{2}-21x}{3}=-\frac{30}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{21}{3}\right)x=-\frac{30}{3}$$

Divide -21 by 3.

$$x^{2}-7x=-\frac{30}{3}$$

Divide -30 by 3.

$$x^{2}-7x=-10$$

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

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-10+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-10+\frac{49}{4}$$

Add -10 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{9}{4}$$

Factor x^{2}-7x+\frac{49}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

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

Add \frac{7}{2} to both sides of the equation.

$$x=5 x=2$$