Solve (z^2+z)+(z^2-11)

$$( { z }^{ 2 } +z)+( { z }^{ 2 } -11)$$

Factor

$$2\left(z-\frac{-\sqrt{89}-1}{4}\right)\left(z-\frac{\sqrt{89}-1}{4}\right)$$

Steps Using the Quadratic Formula

Combine z^{2} and z^{2} to get 2z^{2}.

$$factor(2z^{2}+z-11)$$

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.

$$2z^{2}+z-11=0$$

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.

$$z=\frac{-1±\sqrt{1^{2}-4\times 2\left(-11\right)}}{2\times 2}$$

Square 1.

$$z=\frac{-1±\sqrt{1-4\times 2\left(-11\right)}}{2\times 2}$$

Multiply -4 times 2.

$$z=\frac{-1±\sqrt{1-8\left(-11\right)}}{2\times 2}$$

Multiply -8 times -11.

$$z=\frac{-1±\sqrt{1+88}}{2\times 2}$$

Add 1 to 88.

$$z=\frac{-1±\sqrt{89}}{2\times 2}$$

Multiply 2 times 2.

$$z=\frac{-1±\sqrt{89}}{4}$$

Now solve the equation z=\frac{-1±\sqrt{89}}{4} when ± is plus. Add -1 to \sqrt{89}.

$$z=\frac{\sqrt{89}-1}{4}$$

Now solve the equation z=\frac{-1±\sqrt{89}}{4} when ± is minus. Subtract \sqrt{89} from -1.

$$z=\frac{-\sqrt{89}-1}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1+\sqrt{89}}{4} for x_{1} and \frac{-1-\sqrt{89}}{4} for x_{2}.

$$2z^{2}+z-11=2\left(z-\frac{\sqrt{89}-1}{4}\right)\left(z-\frac{-\sqrt{89}-1}{4}\right)$$

Evaluate

$$2z^{2}+z-11$$

Solution Steps

Combine z^{2} and z^{2} to get 2z^{2}.

$$2z^{2}+z-11$$