## Elementary Algebra

Step 1: $-x^{2}+3x-4=0$ can be rewritten as $x^{2}-3x+4=0$ . Comparing $x^{2}-3x+4=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$; $a=1$, $b=-3$ 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: $x=\frac{-(-3) \pm \sqrt {(-3)^{2}-4(1)(4)}}{2(1)}$ Step 4: $x=\frac{3 \pm \sqrt {9-16}}{2}$ Step 5: $x=\frac{3 \pm \sqrt {-7}}{2}$ Step 6: Since $\sqrt {-7}$ is not a real number, there are no real solutions to the equation. Therefore, the question statement is false.