Answer
No real number solutions.
Work Step by Step
Step 1: Comparing $2x^{2}-3x+7=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find:
$a=2$, $b=-3$ and $c=7$
Step 2: The quadratic formula is:
$x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Step 3: Substituting the values of a, b and c in the formula:
$x=\frac{-(-3) \pm \sqrt {(-3)^{2}-4(2)(7)}}{2(2)}$
Step 4: $x=\frac{3 \pm \sqrt {9-56}}{4}$
Step 5: $x=\frac{3 \pm \sqrt {-47}}{4}$
The square root of -47 is not a real number, so there are no real number solutions.
