Solve {l}{x=frac{2+sqrt{2^2-4(8)(-8)}}{2(6)}}{x_{1}=-1,quadx_{2}=8/6}

$$\left. \begin{array} { l } { x = \frac { 2 + \sqrt { 2 ^ { 2 } - 4 ( 8 ) ( - 8 ) } } { 2 ( 6 ) } } \\ { x _ { 1 } = - 1 , \quad x _ { 2 } = \frac { 8 } { 6 } } \end{array} \right.$$

Solve for x, x_1, x_2

$$x = \frac{\sqrt{65} + 1}{6} \approx 1.510376291$$

$$x_{1}=-1$$

$$x_{2} = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Solution Steps

Consider the first equation. Calculate 2 to the power of 2 and get 4.

$$x=\frac{2+\sqrt{4-4\times 8\left(-8\right)}}{2\times 6}$$

Multiply 4 and 8 to get 32.

$$x=\frac{2+\sqrt{4-32\left(-8\right)}}{2\times 6}$$

Multiply 32 and -8 to get -256.

$$x=\frac{2+\sqrt{4-\left(-256\right)}}{2\times 6}$$

The opposite of -256 is 256.

$$x=\frac{2+\sqrt{4+256}}{2\times 6}$$

Add 4 and 256 to get 260.

$$x=\frac{2+\sqrt{260}}{2\times 6}$$

Factor 260=2^{2}\times 65. Rewrite the square root of the product \sqrt{2^{2}\times 65} as the product of square roots \sqrt{2^{2}}\sqrt{65}. Take the square root of 2^{2}.

$$x=\frac{2+2\sqrt{65}}{2\times 6}$$

Multiply 2 and 6 to get 12.

$$x=\frac{2+2\sqrt{65}}{12}$$

Divide each term of 2+2\sqrt{65} by 12 to get \frac{1}{6}+\frac{1}{6}\sqrt{65}.

$$x=\frac{1}{6}+\frac{1}{6}\sqrt{65}$$

Consider the third equation. Reduce the fraction \frac{8}{6} to lowest terms by extracting and canceling out 2.

$$x_{2}=\frac{4}{3}$$

The system is now solved.

$$x=\frac{1}{6}+\frac{1}{6}\sqrt{65} x_{1}=-1 x_{2}=\frac{4}{3}$$