Answer
{$0,\frac{7}{2}$}
Work Step by Step
Step 1: $2x^{2}=7x$ can also be written as $2x^{2}-7x+0=0$. Comparing $2x^{2}-7x+0=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain:
$a=2$, $b=-7$, and $c=0$
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, we obtain:
$x=\frac{-(-7) \pm \sqrt {(-7)^{2}-4(2)(0)}}{2(2)}$
Step 4: $x=\frac{7 \pm \sqrt {49-0}}{4}$
Step 5: $x=\frac{7 \pm \sqrt {49}}{4}$
Step 6: $x=\frac{7 \pm 7}{4}$
Step 7: $x=\frac{7+7}{4}$ or $x=\frac{7-7}{4}$
Step 8: $x=\frac{14}{4}$ or $x=\frac{0}{4}$
Step 9: $x=\frac{7}{2}$ or $x=0$
Step 10: Therefore, the solution set is {$0,\frac{7}{2}$}.
