What are the **real solutions** of the equation $2x^{2} - x - 6 = 0$?

Question

Accepted Answer

B. $x = 2$ and $x = -\dfrac{3}{2}$

Use the quadratic formula with $a = 2$, $b = -1$, $c = -6$:
$x = \dfrac{-(-1) \pm \sqrt{(-1)^{2} - 4(2)(-6)}}{2(2)} = \dfrac{1 \pm \sqrt{1 + 48}}{4} = \dfrac{1 \pm \sqrt{49}}{4} = \dfrac{1 \pm 7}{4}$.
Two roots: $\dfrac{1 + 7}{4} = 2$ and $\dfrac{1 - 7}{4} = -\dfrac{3}{2}$.
Check by factoring: $2x^{2} - x - 6 = (2x + 3)(x - 2)$. Setting each factor to zero gives $x = -\dfrac{3}{2}$ and $x = 2$ ✓.
Trap A has the signs flipped. Traps C and D do not satisfy the equation — substitute back to verify.

Answer

$x = -2$ and $x = \dfrac{3}{2}$

Answer

$x = 3$ and $x = -2$

Answer

$x = 6$ and $x = -1$