The Quadratic Formula
Key Questions

The quadratic formula is used to get the roots of a quadratic equation, if the roots exists at all.
We usually just perform factorization to get the roots of a quadratic equation. However, this is not always possible (especially when the roots are irrational)
The quadratic formula is
#x = (b + root 2 (b^2  4ac))/(2a)#
Example 1:
#y = x^2 3x  4#
#0 = x^2 3x  4# #=> 0 = (x  4)(x + 1)#
#=> x = 4, x = 1# Using the quadratic formula, let's try to solve the same equation
#x = ((3) + root 2 ((3)^2  4*1*(4)))/(2 * 1)#
#=> x = (3 + root 2 (9 + 16))/2#
#=> x = (3 + root 2 (25))/2#
#=> x = (3 + 5)/2, x = (3  5)/2#
#=> x = 4, x = 1#
Example 2:
#y = 2x^2 3x  5#
#0 = 2x^2  3x  5# Performing factorization is a little hard for this equation, so let's jump straight to using the quadratic formula
#x = ((3) + root 2 ((3)^2  4 * 2 * (5)))/(2 * 2)# #x = (3 + root 2 (9 + 40))/4# #x = (3 + root 2 49)/4# #x = (3 + 7)/4, x = (3  7)/4# #x = 5/2, x = 1# 
Suppose that you have a function represented by
#f(x) = Ax^2 + Bx + C# .We can use the quadratic formula to find the zeroes of this function, by setting
#f(x) = Ax^2 + Bx + C = 0# .Technically we can also find complex roots for it, but typically one will be asked to work only with real roots. The quadratic formula is represented as:
#(B + sqrt(B^24AC))/(2A) = x# ... where x represents the xcoordinate of the zero.
If
#B^2 4AC <0# , we will be dealing with complex roots, and if#B^2  4AC >=0# , we will have real roots.As an example, consider the function
#x^2 13x + 12# . Here,#A = 1, B = 13, C = 12.# Then for the quadratic formula we would have:
# x = (13 + sqrt ((13)^2  4(1)(12)))/(2(1))# =#(13 + sqrt (169  48))/2 = (13+11)/2# Thus, our roots are
#x=1# and#x=12# .For an example with complex roots, we have the function
#f(x) =x^2 +1# . Here#A = 1, B = 0, C = 1.# Then by the quadratic equation,
#x = (0 + sqrt (0^2  4(1)(1)))/(2(1)) = +sqrt(4)/2 = +i# ... where
#i# is the imaginary unit, defined by its property of#i^2 = 1# .In the graph for this function on the real coordinate plane, we will see no zeroes, but the function will have these two imaginary roots.

The discriminant is part of the quadratic formula.
Quadratic Formula
#x=(b+sqrt(b^24ac))/(2a)# Discriminant
#b^24ac# The discriminant tells you the number and types of solutions to a quadratic equation.
#b^24ac = 0# , one real solution#b^24ac > 0# , two real solutions#b^24ac < 0# , two imaginary solutions 
Answer:
#x=(b+sqrt(b^24ac))/(2a)# Explanation:
Negative b plus minus the square root of b squared minus 4*a*c over 2*a. To plug something into the quadratic formula the equation needs to be in standard form (
#ax^2 + bx^2 +c # ).hope this helps!