Solve 15x(x-1)=30

$$15 x ( x - 1 ) = 30$$

Solve for x

$$x=2$$

$$x=-1$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 15x by x-1.

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

Subtract 30 from both sides.

$$15x^{2}-15x-30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 15 for a, -15 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 15\left(-30\right)}}{2\times 15}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 15\left(-30\right)}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-\left(-15\right)±\sqrt{225-60\left(-30\right)}}{2\times 15}$$

Multiply -60 times -30.

$$x=\frac{-\left(-15\right)±\sqrt{225+1800}}{2\times 15}$$

Add 225 to 1800.

$$x=\frac{-\left(-15\right)±\sqrt{2025}}{2\times 15}$$

Take the square root of 2025.

$$x=\frac{-\left(-15\right)±45}{2\times 15}$$

The opposite of -15 is 15.

$$x=\frac{15±45}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{15±45}{30}$$

Now solve the equation x=\frac{15±45}{30} when ± is plus. Add 15 to 45.

$$x=\frac{60}{30}$$

Divide 60 by 30.

$$x=2$$

Now solve the equation x=\frac{15±45}{30} when ± is minus. Subtract 45 from 15.

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

Divide -30 by 30.

$$x=-1$$

The equation is now solved.

$$x=2 x=-1$$

Steps for Completing the Square

Use the distributive property to multiply 15x by x-1.

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

Divide both sides by 15.

$$\frac{15x^{2}-15x}{15}=\frac{30}{15}$$

Dividing by 15 undoes the multiplication by 15.

$$x^{2}+\left(-\frac{15}{15}\right)x=\frac{30}{15}$$

Divide -15 by 15.

$$x^{2}-x=\frac{30}{15}$$

Divide 30 by 15.

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

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

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=2+\frac{1}{4}$$

Add 2 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{9}{4}$$

Factor x^{2}-x+\frac{1}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{1}{2}=\frac{3}{2} x-\frac{1}{2}=-\frac{3}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=2 x=-1$$