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solve for the solutions of $3x^2 + 2x + 8 = 0$. use the keypad to enter…

Question

solve for the solutions of $3x^2 + 2x + 8 = 0$. use the keypad to enter your answers in the boxes. additional symbols can be found using the drop - down arrows at the top of the keypad. the solutions are $x = \square$ and $x = \square$.

Explanation:

Step1: Identify quadratic coefficients

For $ax^2+bx+c=0$, here $a=3$, $b=2$, $c=8$

Step2: Calculate discriminant

$\Delta = b^2-4ac = 2^2 - 4(3)(8) = 4 - 96 = -92$

Step3: Apply quadratic formula

$x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-2 \pm \sqrt{-92}}{6}$

Step4: Simplify complex root

$\sqrt{-92} = \sqrt{4 \times (-23)} = 2i\sqrt{23}$, so $x = \frac{-2 \pm 2i\sqrt{23}}{6} = \frac{-1 \pm i\sqrt{23}}{3}$

Answer:

$x = \frac{-1 + i\sqrt{23}}{3}$ and $x = \frac{-1 - i\sqrt{23}}{3}$