Question

question 4 of 10
the quadratic formula gives which roots for the equation $3x^2 + 3x = 2$
a. $x = \frac{-3\pm\sqrt{33}}{4}$
b. $x = \frac{-3\pm\sqrt{33}}{6}$
c. $x = \frac{-3\pm\sqrt{35}}{6}$
d. $x = \frac{3\pm\sqrt{35}}{3}$

Explanation:

Step1: Rewrite to standard form

$3x^2 + 3x - 2 = 0$ (so $a=3$, $b=3$, $c=-2$)

Step2: Calculate discriminant

$\Delta = b^2 - 4ac = 3^2 - 4(3)(-2) = 9 + 24 = 33$

Step3: Apply quadratic formula

$x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-3 \pm \sqrt{33}}{2(3)} = \frac{-3 \pm \sqrt{33}}{6}$

Answer:

B. $x = \frac{-3\pm\sqrt{33}}{6}$