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

answer:   14

step-by-step explanation:          

pu you stank bo u ugly bo lol

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)