To solve the quadratic equation \(8x^2 + 5 = 12x\), we first rearrange it into standard form by moving all terms to one side:
\[ 8x^2 - 12x + 5 = 0 \]
Next, we identify the coefficients \(a\), \(b\), and \(c\):
- \(a = 8\)
- \(b = -12\)
- \(c = 5\)
We will use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Now, we calculate the discriminant \(b^2 - 4ac\):
\[ b^2 = (-12)^2 = 144 \] \[ 4ac = 4 \cdot 8 \cdot 5 = 160 \] \[ b^2 - 4ac = 144 - 160 = -16 \]
Since the discriminant is negative, we will have complex solutions.
Now, substituting the values into the quadratic formula:
\[ x = \frac{-(-12) \pm \sqrt{-16}}{2 \cdot 8} \] \[ x = \frac{12 \pm \sqrt{-16}}{16} \] \[ x = \frac{12 \pm 4i}{16} \] \[ x = \frac{12}{16} \pm \frac{4i}{16} \] \[ x = \frac{3}{4} \pm \frac{1}{4}i \]
Thus, the solutions are:
\[ x = \frac{3}{4} \pm \frac{1}{4}i \]
Now we compare this with the given options:
A: \(\frac{3}{4} \pm \frac{1}{4}i\) (which is correct) B: \(12 \pm 4i\) C: \(12 \pm \frac{1}{4}i\) D: \(-\frac{3}{4} + \frac{1}{4}i\)
The correct answer is:
A: \(\frac{3}{4} \pm \frac{1}{4}i\)