Solving quadratic equations — middle term splitting and quadratic formula

A quadratic equation has the form $a{x}^2+bx+c=0$ where $a\ne 0$ and the highest power of the variable is exactly 2.

The choice “$x^3+2x^2-5=0$” has degree 3 (cubic). The choice “$x^2-4x+7=0$” has degree 2. The choice “$3x+7=0$” has degree 1 (linear). The choice “$\frac{1}{x^2}+3=0$” involves $\frac{1}{x^2}$, which is not a polynomial.

$x^2-4x+7=0$ is the only polynomial of degree exactly 2, so it is the quadratic equation.

$x^3+2x^2-5=0$ contains $x^3$, making it cubic (degree 3), not quadratic.

$3x+7=0$ has highest power 1, making it a linear equation.

$\frac{1}{x^2}+3=0$ is not a polynomial equation — it cannot be written as $a{x}^2+bx+c=0$.

Only $x^2-4x+7=0$ satisfies all conditions of a quadratic equation.

$x=\pm \sqrt{81}$

$\sqrt{81}=9$, so $x=\pm 9$.

$9^2=81$ ✓ and $(-9{)}^2=81$ ✓. Both values work.

$x=9$ only misses the negative solution. Since $(-9{)}^2=81$, this is incomplete.

$x=-9$ only misses the positive solution. Since $9^2=81$, this is also incomplete.

$\pm 40.5$ is incorrect — $\sqrt{81}=9$, not $\frac{81}{2}$.

$x=\pm 9$ gives both solutions, which is always required when taking a square root.