Solve (4)13z-6z^2-6

$$( 4 ) 13 z - 6 z ^ { 2 } - 6$$

Factor

$$-6\left(z-\frac{13-4\sqrt{10}}{3}\right)\left(z-\frac{4\sqrt{10}+13}{3}\right)$$

Steps Using the Quadratic Formula

Multiply 4 and 13 to get 52.

$$factor(52z-6z^{2}-6)$$

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.

$$-6z^{2}+52z-6=0$$

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.

$$z=\frac{-52±\sqrt{52^{2}-4\left(-6\right)\left(-6\right)}}{2\left(-6\right)}$$

Square 52.

$$z=\frac{-52±\sqrt{2704-4\left(-6\right)\left(-6\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$z=\frac{-52±\sqrt{2704+24\left(-6\right)}}{2\left(-6\right)}$$

Multiply 24 times -6.

$$z=\frac{-52±\sqrt{2704-144}}{2\left(-6\right)}$$

Add 2704 to -144.

$$z=\frac{-52±\sqrt{2560}}{2\left(-6\right)}$$

Take the square root of 2560.

$$z=\frac{-52±16\sqrt{10}}{2\left(-6\right)}$$

Multiply 2 times -6.

$$z=\frac{-52±16\sqrt{10}}{-12}$$

Now solve the equation z=\frac{-52±16\sqrt{10}}{-12} when ± is plus. Add -52 to 16\sqrt{10}.

$$z=\frac{16\sqrt{10}-52}{-12}$$

Divide -52+16\sqrt{10} by -12.

$$z=\frac{13-4\sqrt{10}}{3}$$

Now solve the equation z=\frac{-52±16\sqrt{10}}{-12} when ± is minus. Subtract 16\sqrt{10} from -52.

$$z=\frac{-16\sqrt{10}-52}{-12}$$

Divide -52-16\sqrt{10} by -12.

$$z=\frac{4\sqrt{10}+13}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{13-4\sqrt{10}}{3} for x_{1} and \frac{13+4\sqrt{10}}{3} for x_{2}.

$$-6z^{2}+52z-6=-6\left(z-\frac{13-4\sqrt{10}}{3}\right)\left(z-\frac{4\sqrt{10}+13}{3}\right)$$

Evaluate

$$-6z^{2}+52z-6$$

Solution Steps

Multiply 4 and 13 to get 52.

$$52z-6z^{2}-6$$