Solve (3z+1)(z-4)=-5(z-1)

$$( 3 z + 1 ) ( z - 4 ) = - 5 ( z - 1 )$$

Solve for z

$$z=-1$$

$$z=3$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3z+1 by z-4 and combine like terms.

$$3z^{2}-11z-4=-5\left(z-1\right)$$

Use the distributive property to multiply -5 by z-1.

$$3z^{2}-11z-4=-5z+5$$

Add 5z to both sides.

$$3z^{2}-11z-4+5z=5$$

Combine -11z and 5z to get -6z.

$$3z^{2}-6z-4=5$$

Subtract 5 from both sides.

$$3z^{2}-6z-4-5=0$$

Subtract 5 from -4 to get -9.

$$3z^{2}-6z-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -6 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 3\left(-9\right)}}{2\times 3}$$

Square -6.

$$z=\frac{-\left(-6\right)±\sqrt{36-4\times 3\left(-9\right)}}{2\times 3}$$

Multiply -4 times 3.

$$z=\frac{-\left(-6\right)±\sqrt{36-12\left(-9\right)}}{2\times 3}$$

Multiply -12 times -9.

$$z=\frac{-\left(-6\right)±\sqrt{36+108}}{2\times 3}$$

Add 36 to 108.

$$z=\frac{-\left(-6\right)±\sqrt{144}}{2\times 3}$$

Take the square root of 144.

$$z=\frac{-\left(-6\right)±12}{2\times 3}$$

The opposite of -6 is 6.

$$z=\frac{6±12}{2\times 3}$$

Multiply 2 times 3.

$$z=\frac{6±12}{6}$$

Now solve the equation z=\frac{6±12}{6} when ± is plus. Add 6 to 12.

$$z=\frac{18}{6}$$

Divide 18 by 6.

$$z=3$$

Now solve the equation z=\frac{6±12}{6} when ± is minus. Subtract 12 from 6.

$$z=-\frac{6}{6}$$

Divide -6 by 6.

$$z=-1$$

The equation is now solved.

$$z=3 z=-1$$

Steps for Completing the Square

Use the distributive property to multiply 3z+1 by z-4 and combine like terms.

$$3z^{2}-11z-4=-5\left(z-1\right)$$

Use the distributive property to multiply -5 by z-1.

$$3z^{2}-11z-4=-5z+5$$

Add 5z to both sides.

$$3z^{2}-11z-4+5z=5$$

Combine -11z and 5z to get -6z.

$$3z^{2}-6z-4=5$$

Add 4 to both sides.

$$3z^{2}-6z=5+4$$

Add 5 and 4 to get 9.

$$3z^{2}-6z=9$$

Divide both sides by 3.

$$\frac{3z^{2}-6z}{3}=\frac{9}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$z^{2}+\left(-\frac{6}{3}\right)z=\frac{9}{3}$$

Divide -6 by 3.

$$z^{2}-2z=\frac{9}{3}$$

Divide 9 by 3.

$$z^{2}-2z=3$$

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

$$z^{2}-2z+1=3+1$$

Add 3 to 1.

$$z^{2}-2z+1=4$$

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$z-1=2 z-1=-2$$

Add 1 to both sides of the equation.

$$z=3 z=-1$$