Solve the equation $2x^2 - 7x - 3 = 0$ - Leaving Cert Mathematics - Question 3 - 2018

To solve this quadratic equation, we will use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 2$, $b = -7$, and $c = -3$. Plugging in these values gives:

$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2}$

Calculating inside the square root:

$x = \frac{7 \pm \sqrt{49 + 24}}{4} = \frac{7 \pm \sqrt{73}}{4}$

Now compute the two possible values for $x$:

1. $x_1 = \frac{7 + \sqrt{73}}{4} \approx 3.89$
2. $x_2 = \frac{7 - \sqrt{73}}{4} \approx -0.39$

Therefore, the solutions to the equation are approximately $x \approx 3.89$ and $x \approx -0.39$.