Question $\newline$ Solve the equation for all real solutions in simplest form. $\newline$ $3 z^{2}-z+5=6$ $\newline$ Answer Attempt $1$ out of $2$ $\newline$ Additional Solution $\newline$ No Solution $\newline$ $z=$ $\newline$ Submit Answer

Full solution

Q. Question

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Solve the equation for all real solutions in simplest form.

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3 z^{2}-z+5=6

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Answer Attempt

1

out of

2

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Additional Solution

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No Solution

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z=

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Submit Answer

Set Equation to Zero: First, we need to set the equation to zero by subtracting $6$ from both sides of the equation. $\newline$ $3z^2 - z + 5 = 6$ $\newline$ $3z^2 - z + 5 - 6 = 0$ $\newline$ $3z^2 - z - 1 = 0$
Use Quadratic Formula: Now, we attempt to factor the quadratic equation, but it does not factor nicely. Therefore, we will use the quadratic formula to find the solutions for $z$ . The quadratic formula is $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 3$ , $b = -1$ , and $c = -1$ .
Calculate Discriminant: We calculate the discriminant $b^2 - 4ac$ to determine the number of real solutions. $\newline$ Discriminant = $(-1)^2 - 4(3)(-1)$ $\newline$ Discriminant = $1 + 12$ $\newline$ Discriminant = $13$ $\newline$ Since the discriminant is positive, there are two real solutions.
Substitute Values: We substitute the values of $a$ , $b$ , and $c$ into the quadratic formula to find the solutions for $z$ . $\newline$ $z = \frac{-(-1) \pm \sqrt{(13)}}{(2 \cdot 3)}$ $\newline$ $z = \frac{(1 \pm \sqrt{13})}{6}$
Find Solutions: We now have the two solutions for $z$ in simplest form: $z = \frac{1 + \sqrt{13}}{6}$ and $z = \frac{1 - \sqrt{13}}{6}$