Answer
No solution within the realm of real numbers. When imaginary numbers are included:
$x = \frac{2\frac{+}{} \sqrt {3}i}{2}$
Work Step by Step
$$4x^{2} = 8x -7$$
$4x^{2} - 8x + 7 = 0$
where we can solve this quadratic function by finding the zeros using the quadratic formula $\frac{-b\frac{+}{} \sqrt {b^{2} - 4ac}}{2a}$:
$x = \frac{-(-8)\frac{+}{} \sqrt {(-8)^{2} - 4(4)(7)}}{2(4)} = \frac{8\frac{+}{} \sqrt {(64 - 112}}{8} = \frac{8\frac{+}{} \sqrt {-48}}{8}$
Since $\sqrt {-48}$ is not a Real number, we can say that this equation does not have any solutions within the realm of Real numbers. If we were to include Imaginary numbers, the solutions would result in the following:
$x = \frac{8\frac{+}{} \sqrt {-48}}{8} = \frac{8\frac{+}{} \sqrt {48}i}{8} = \frac{8\frac{+}{} \sqrt {(16)(3)}i}{8} = \frac{8\frac{+}{} 4\sqrt {3}i}{8}$
where $x = \frac{2\frac{+}{} \sqrt {3}i}{2}$
