Solve z+z^2-1

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

Factor

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

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$z^{2}+z-1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

Square 1.

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

Multiply -4 times -1.

$$z=\frac{-1±\sqrt{1+4}}{2}$$

Add 1 to 4.

$$z=\frac{-1±\sqrt{5}}{2}$$

Now solve the equation z=\frac{-1±\sqrt{5}}{2} when ± is plus. Add -1 to \sqrt{5}.

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

Now solve the equation z=\frac{-1±\sqrt{5}}{2} when ± is minus. Subtract \sqrt{5} from -1.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1+\sqrt{5}}{2} for x_{1} and \frac{-1-\sqrt{5}}{2} for x_{2}.

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

Evaluate

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