Any quadratic curve can be written in the form: $$ \begin{equation}\begin{aligned} ax^2+bx+c\\ \end{aligned}\end{equation} $$
We know that equations of the form $$ \begin{equation}\begin{aligned} ax^2+bx+c=0\\ \end{aligned}\end{equation} $$
can be solved using the quadratic formula: $$ \begin{equation}\begin{aligned} x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \end{aligned}\end{equation} $$
In some cases, by using factorization alongside the zero factor property.
The zero factor property dictates that if a product of two values is zero ($ab$=0) then either of the two values must be zero ($a=0$ or $b=0$).
This is all to say that if we have a quadratic equation, we have set ways in which we can solve them.
Some equations are quadratic without actually looking quadratic
Consider the equation: $$ \begin{equation}\begin{aligned} x^4-5x^2+6=0\\ \end{aligned}\end{equation} $$
You might be tempted to treat this as a quartic expression and maybe use a graphing calculator to find its roots. If, however, we let $y=x^2$, we get a familiar sight: $$ \begin{equation}\begin{aligned} y^2-5y+6=0\\ \end{aligned}\end{equation} $$
This is a quadratic equation. Thus, we can solve it via factorization: $$ \begin{equation}\begin{aligned} y^2-5y+6&=0\\ y^2-3y-2y+6&=0\\ (y^2-3y)-(2y-6)&=0\\ y(y-3)-2(y-3)&=0\\ (y-3)(y-2)&=0\\ \end{aligned}\end{equation} $$
Applying the zero factor property: $$ \begin{equation}\begin{aligned} y-3=0\\ \therefore y=3\\ y-2=0\\ \therefore y=2\\ \end{aligned}\end{equation} $$
These $y$ solve the equation $y^2-5x+6$ but we still need to find the corresponding $x$ values which satisfy the equation $x^4-5x+6$. We do this by substituting $y$ with its original value $x^2$: $$ \begin{equation}\begin{aligned} x^2&=3\\ \therefore x&=\pm\sqrt{3}\\ x^2&=2\\ \therefore x&=\pm\sqrt{2}\\ \end{aligned}\end{equation} $$
Thus the solutions for the original equation are $\sqrt{3}$, $-\sqrt{3}$, $\sqrt{2}$ and $-\sqrt{2}$.
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