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
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Answer 1

x = 5 and x = -19

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