Solve 5z(z-5)=-30

$$5 z ( z - 5 ) = - 30$$

Solve for z

$$z=2$$

$$z=3$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 5z by z-5.

$$5z^{2}-25z=-30$$

Add 30 to both sides.

$$5z^{2}-25z+30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -25 for b, and 30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 5\times 30}}{2\times 5}$$

Square -25.

$$z=\frac{-\left(-25\right)±\sqrt{625-4\times 5\times 30}}{2\times 5}$$

Multiply -4 times 5.

$$z=\frac{-\left(-25\right)±\sqrt{625-20\times 30}}{2\times 5}$$

Multiply -20 times 30.

$$z=\frac{-\left(-25\right)±\sqrt{625-600}}{2\times 5}$$

Add 625 to -600.

$$z=\frac{-\left(-25\right)±\sqrt{25}}{2\times 5}$$

Take the square root of 25.

$$z=\frac{-\left(-25\right)±5}{2\times 5}$$

The opposite of -25 is 25.

$$z=\frac{25±5}{2\times 5}$$

Multiply 2 times 5.

$$z=\frac{25±5}{10}$$

Now solve the equation z=\frac{25±5}{10} when ± is plus. Add 25 to 5.

$$z=\frac{30}{10}$$

Divide 30 by 10.

$$z=3$$

Now solve the equation z=\frac{25±5}{10} when ± is minus. Subtract 5 from 25.

$$z=\frac{20}{10}$$

Divide 20 by 10.

$$z=2$$

The equation is now solved.

$$z=3 z=2$$

Steps for Completing the Square

Use the distributive property to multiply 5z by z-5.

$$5z^{2}-25z=-30$$

Divide both sides by 5.

$$\frac{5z^{2}-25z}{5}=-\frac{30}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$z^{2}+\left(-\frac{25}{5}\right)z=-\frac{30}{5}$$

Divide -25 by 5.

$$z^{2}-5z=-\frac{30}{5}$$

Divide -30 by 5.

$$z^{2}-5z=-6$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$z^{2}-5z+\left(-\frac{5}{2}\right)^{2}=-6+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$z^{2}-5z+\frac{25}{4}=-6+\frac{25}{4}$$

Add -6 to \frac{25}{4}.

$$z^{2}-5z+\frac{25}{4}=\frac{1}{4}$$

Factor z^{2}-5z+\frac{25}{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}.

$$\left(z-\frac{5}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z-\frac{5}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$z-\frac{5}{2}=\frac{1}{2} z-\frac{5}{2}=-\frac{1}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$z=3 z=2$$