Solve 6z^2-11z=-7z-4

$$6 z ^ { 2 } - 11 z = - 7 z - 4$$

Solve for z

$$z=\frac{1+\sqrt{5}i}{3}\approx 0.333333333+0.745355992i$$

$$z=\frac{-\sqrt{5}i+1}{3}\approx 0.333333333-0.745355992i$$

Steps Using the Quadratic Formula

Add 7z to both sides.

$$6z^{2}-11z+7z=-4$$

Combine -11z and 7z to get -4z.

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

Add 4 to both sides.

$$6z^{2}-4z+4=0$$

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

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

Square -4.

$$z=\frac{-\left(-4\right)±\sqrt{16-4\times 6\times 4}}{2\times 6}$$

Multiply -4 times 6.

$$z=\frac{-\left(-4\right)±\sqrt{16-24\times 4}}{2\times 6}$$

Multiply -24 times 4.

$$z=\frac{-\left(-4\right)±\sqrt{16-96}}{2\times 6}$$

Add 16 to -96.

$$z=\frac{-\left(-4\right)±\sqrt{-80}}{2\times 6}$$

Take the square root of -80.

$$z=\frac{-\left(-4\right)±4\sqrt{5}i}{2\times 6}$$

The opposite of -4 is 4.

$$z=\frac{4±4\sqrt{5}i}{2\times 6}$$

Multiply 2 times 6.

$$z=\frac{4±4\sqrt{5}i}{12}$$

Now solve the equation z=\frac{4±4\sqrt{5}i}{12} when ± is plus. Add 4 to 4i\sqrt{5}.

$$z=\frac{4+4\sqrt{5}i}{12}$$

Divide 4+4i\sqrt{5} by 12.

$$z=\frac{1+\sqrt{5}i}{3}$$

Now solve the equation z=\frac{4±4\sqrt{5}i}{12} when ± is minus. Subtract 4i\sqrt{5} from 4.

$$z=\frac{-4\sqrt{5}i+4}{12}$$

Divide 4-4i\sqrt{5} by 12.

$$z=\frac{-\sqrt{5}i+1}{3}$$

The equation is now solved.

$$z=\frac{1+\sqrt{5}i}{3} z=\frac{-\sqrt{5}i+1}{3}$$

Steps for Completing the Square

Add 7z to both sides.

$$6z^{2}-11z+7z=-4$$

Combine -11z and 7z to get -4z.

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

Divide both sides by 6.

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

Dividing by 6 undoes the multiplication by 6.

$$z^{2}+\left(-\frac{4}{6}\right)z=-\frac{4}{6}$$

Reduce the fraction \frac{-4}{6} to lowest terms by extracting and canceling out 2.

$$z^{2}-\frac{2}{3}z=-\frac{4}{6}$$

Reduce the fraction \frac{-4}{6} to lowest terms by extracting and canceling out 2.

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

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$z^{2}-\frac{2}{3}z+\left(-\frac{1}{3}\right)^{2}=-\frac{2}{3}+\left(-\frac{1}{3}\right)^{2}$$

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$z^{2}-\frac{2}{3}z+\frac{1}{9}=-\frac{2}{3}+\frac{1}{9}$$

Add -\frac{2}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$z^{2}-\frac{2}{3}z+\frac{1}{9}=-\frac{5}{9}$$

Factor z^{2}-\frac{2}{3}z+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(z-\frac{1}{3}\right)^{2}=-\frac{5}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z-\frac{1}{3}\right)^{2}}=\sqrt{-\frac{5}{9}}$$

Simplify.

$$z-\frac{1}{3}=\frac{\sqrt{5}i}{3} z-\frac{1}{3}=-\frac{\sqrt{5}i}{3}$$

Add \frac{1}{3} to both sides of the equation.

$$z=\frac{1+\sqrt{5}i}{3} z=\frac{-\sqrt{5}i+1}{3}$$