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

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

let y equal to

the total fare and x is the mile of taxi ride. so the equation is

y = 0.55x + 1.75

since susie has

$10 to spend for a taxi cab, so he can have

10 = 0.55x +

1.75

x = 15 miles of

taxi ride

so the system of

inequality is

10 < 0.55x +

1.75

x > 2

the revenue of the fitness center, y, is the product of two factors: the number of members and the fee paid by each member. there are 320 members now, but number of members goes down by 10 for each $5 increase in fees, so if x is the number of $5 price increases then members is equal to (320 - 10x). since the fee is currently $45, and the fee increases by $5 for every price bump x, the fee is given by (45 + 5x). so, revenue y is given by y = (320 - 10x)(45 + 5x).