## Elementary Algebra

The only way to check if the equation has two real number solutions or not is to solve the equation: Step 1: Comparing $5x^{2}-x+4=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find: $a=5$, $b=-1$ and $c=4$ 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: $x=\frac{-(-1) \pm \sqrt {(-1)^{2}-4(5)(4)}}{2(5)}$ Step 4: $x=\frac{1 \pm \sqrt {1-80}}{10}$ Step 5: $x=\frac{1 \pm \sqrt {-79}}{10}$ $\sqrt{-79}$ is a non-real number. Therefore, without even simplifying, we can see that the solutions to the equation will not be real numbers, making the statement false.