Question

How do I find if this type of equation has one solution two solutions or no solution 5x^2+14=19 I tried using the b^-4ac method but I don't know what numbers to plug in for b and c
discriminant
​

Answer 1

Answer:

x = 5 and x = -19

Step-by-step explanation:

You're on the right track.  It's the "discriminant" that tells you what you want to know here.  Before starting, arrange the terms of your quadratic in descending orders of x:  5x^2 + 14x - 19 = 0 (Note that I assumed you meant 14x instead of just 14).

Then the coefficients of this quadratic are a = 5, b = 14 and c = -19.

You are referring to the "quadratic formula."  It states this:

      -b ± √(b²-4ac)

x = -----------------------

               2a                                        

So, we insert the a, b and c values as indicated above:

      -14 ± √( 14² - 4[5][-19] )        -14 ± √(196 - 4[5][-19] )        -14 ± √576

x = ----------------------------------- = ---------------------------------- = ----------------------

                 2(10)                                 20                                      20

This comes out to:

x = (-14 + 24) / 2      and     x = (-14 - 24) / 2

or:

x = 5 and x = -19