How do you solve #y = x^2 + 3x + 2#?

Question

Nathan Axel · Accepted Answer

It depends on what you want to solve for.
#color(white)("XXXX")#the y intercept #= 2#
#color(white)("XXXX")#the x intercepts #=-1# and #-2#
#color(white)("XXXX")#the vertex = #(-3/2,1/4)# You are given the generalized solution for #y#. If you want a different solution then what kind of solution you want needs to be specified.

Savannah Anderson · Answer

undefined To solve the equation $y = x^2 + 3x + 2$, you can set $y$ equal to zero and then apply the quadratic formula or factor the quadratic expression.

For the quadratic formula method:
$$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$$

Where $a = 1$, $b = 3$, and $c = 2$. Substitute these values into the formula to find the solutions for $x$.

For factoring method, you factor the quadratic expression $x^2 + 3x + 2$ into two binomials: $(x + ?)(x + ?)$ where the question marks are the numbers that multiply to give you 2 and add to give you 3. Then solve for $x$ by setting each factor equal to zero and solving for $x$.