How do you solve $y = 3{x^2} - x - 2$, $y = - 2x + 2$ using substitution?

Hint: This problem deals with solving a quadratic equation, through the method of substitution from two given expressions. Here, given a quadratic equation expression, we have to simplify the expression and make it into a standard form of quadratic equation. If the quadratic equation is in the form of $a{x^2} + bx + c = 0$, then we know that the roots of this quadratic equation are given by :
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step-by-step answer:
Here the given two expressions are one is a quadratic expression, whereas the other equation is just a linear equation, which are given below:
The first equation is given by:
$ \Rightarrow y = 3{x^2} - x - 2$
Whereas the second question is given by:
$ \Rightarrow y = - 2x + 2$
Both the equations are equated to the same variable which is equal to $y$.
So the expression of $y$ of the second equation is substituted in the first equation in place of $y$, as shown below:
$ \Rightarrow 3{x^2} - x - 2 = - 2x + 2$
Now rearrange the above equation such that all the terms are on one side of the equation as shown:
$ \Rightarrow 3{x^2} - x + 2x - 2 - 2 = 0$
$ \Rightarrow 3{x^2} + x - 4 = 0$
Now the above equation is in the standard form of a quadratic equation, which is $a{x^2} + bx + c = 0$.
Here comparing the equation $3{x^2} + x - 4 = 0$ with the standard form $a{x^2} + bx + c = 0$ and compare the coefficients $a,b$ and $c$:
$ \Rightarrow a = 3,b = 1$ and $c = - 4$
Now applying the formula to find the value of the roots of $x$, as given below:
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Substituting the values of $a,b$ and $c$ in the above formula:
$ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {{{\left( 1 \right)}^2} - 4\left( 3 \right)\left( { - 4} \right)} }}{{2(3)}}$
$ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 + 48} }}{6}$
Simplifying the above expression, as given below:
$ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {49} }}{6}$
We know that the square root of 49 is 7, $\sqrt {49} = 7$
$ \Rightarrow x = \dfrac{{ - 1 \pm 7}}{6}$
Now considering the two cases, with plus and minus, as shown:
$ \Rightarrow x = \dfrac{{ - 1 + 7}}{6};x = \dfrac{{ - 1 - 7}}{6}$
$ \Rightarrow x = \dfrac{6}{2};x = \dfrac{{ - 8}}{2}$
Hence the value of the roots are equal to :
$\therefore x = 3;x = - 4$

Final Answer: The solution of the equations $y = 3{x^2} - x - 2$, $y = - 2x + 2$ using substitution are 3 and -4.

Note:
Please note that this problem can also be done either by the method of completing the square or just factoring and solving the quadratic equation. To solve $a{x^2} + bx + c = 0$ by completing the square: transform the equation so that the constant term,$c$ is alone on the right side. But here we are adding and subtracting some terms in order to factor.