Solving Quadratic Systems

We want to solve the given system of equations using the substitution method. $$\{\begin{array}{ll}2y=y-x^2+1& \text{(I)}\\ y=x^2-5x-2& \text{(II)}\end{array}$$ The $y$-variable is isolated in Equation (II). This allows us to substitute its value $x^2-5x-2$ for $y$ in Equation (I). Notice that in Equation (I), we have a quadratic equation in terms of only the $x$-variable. $$\begin{array}{c}2x^2-5x-3=0\iff 2x^2+(\text{-}5)x+(\text{-}3)=0\end{array}$$ Now, recall the Quadratic Formula. $$\begin{array}{c}x=\frac{\text{-}b\pm \sqrt{b^2-4ac}}{2a}\end{array}$$ We can substitute $a=2,$ $b=\text{-}5,$ and $c=\text{-}3$ into this formula to solve the quadratic equation. This result tells us that we have two solutions for $x.$ One of them will use the positive sign and the other one will use the negative sign.

$x=\frac{5\pm 7}{4}$
$x_1=\frac{5+7}{4}$	$x_2=\frac{5-7}{4}$
$x_1=\frac{12}{4}$	$x_2=\frac{\text{-}2}{4}$
$x_1=3$	$x_2=\text{-}\frac{1}{2}$

Now, consider Equation (II). $$\begin{array}{c}y=x^2-5x-2\end{array}$$ We can substitute $x=3$ and $x=\text{-}\frac{1}{2}$ into the above equation to find the values for $y.$ Let's start with $x=3.$ We found that $y=\text{-}8,$ when $x=3.$ One solution of the system, which is a point of intersection of the two parabolas, is $(3,\text{-}8).$ To find the other solution, we will substitute $\text{-}\frac{1}{2}$ for $x$ in Equation (II) again. We found that $y=\frac{3}{4},$ when $x=\text{-}\frac{1}{2}.$ Therefore, our second solution, which is the other point of intersection of the two parabolas, is $\left(\text{-}\frac{1}{2},\frac{3}{4}\right).$