Answer
Add $\frac{9}{16}$ to both sides of the equation to complete the square.
The solutions are $-\frac{3+\sqrt 41}{4}$ and $-\frac{3-\sqrt 41}{4}$, and the solution set is {$-\frac{3+\sqrt 41}{4},-\frac{3-\sqrt 41}{4}$}
Work Step by Step
$2x^2 + 3x - 4 = 0$
Divide the whole equation by $2$. Thus, the equation becomes:
$x^2 + \frac{3}{2}x - 2 = 0$
Add $2$ from both sides to isolate the binomial $x^2 + \frac{3}{2}x$.
$x^2 + \frac{3}{2}x - 2 +2 = 0+2$
$x^2 + \frac{3}{2}x = 2$
The coefficient of the x-term is $\frac{3}{2}$. Half of $\frac{3}{2}$ is $\frac{3}{4}$, and $(\frac{3}{4})^2 = \frac{9}{16}$.
Thus, add $\frac{9}{16}$ to both sides of the equation to complete the square.
$x^2 + \frac{3}{2}x + \frac{9}{16}= 2+\frac{9}{16}$
$x^2 + \frac{3}{2}x + \frac{9}{16}= \frac{41}{16}$
$(x+\frac{3}{4})^2 = \frac{41}{16}$
$x+\frac{3}{4} = \sqrt \frac{41}{16}$ or $x+\frac{3}{4} = -\sqrt \frac{41}{16}$
$x = - \frac{3}{4} +\sqrt \frac{41}{16}$ or $x = - \frac{3}{4} -\sqrt \frac{41}{16}$
Rationalize the denominator:
$x = - \frac{3}{4} +\frac{\sqrt 41}{4}$ or $x = - \frac{3}{4} -\frac{\sqrt 41}{4}$
$x = -\frac{3+\sqrt 41}{4}$ or $x = -\frac{3-\sqrt 41}{4}$
The solutions are $-\frac{3+\sqrt 41}{4}$ and $-\frac{3-\sqrt 41}{4}$, and the solution set is {$-\frac{3+\sqrt 41}{4},-\frac{3-\sqrt 41}{4}$}.
