Solving using the quadratic formula

Question

Solving using the quadratic formula
72 =3p^2+50p

Answer:
72-(3p^2+50p)=3p^2+50p-(3p^2+50p)
-3p^2-50p+72=0
(-3p+4)(p+18)=0
-3p+4=0 or p+18=0
p=-18

 

I really don't think I did this right someone help please 

 

Answer 1

Hi Jessica, the quadratic formula for the variable p is:

 

p = (-b ±√(b² - 4 * a * c)) / (2 * a)

 

The first step is to get all terms on one side of the equal sign and 0 on the other:

 

3p² + 50p - 72 = 0

 

Substitute for a, b, and c:

 

a = 3

 

b = 50

 

c = -72

 

p = [-50 ±√(50² - (4 * 3 * -72))] / (2 * 3)

 

Solve for p

 

p = [-50 ±√(2500 + 864)] / 6

 

p = [-50 ±√(3364)] / 6

 

p = (-50 ± 58) / 6

 

P = (-50 + 58) / 6, or 1.33

 

and 

 

P = (-50 - 58) / 6, or -18

 

Questions?

 

Answer 2

Jessica, you did this correctly. The only thing is that it says to use the quadratic formula.

(-50±√3364)/6

(-50±58)/6

8/6=4/3

-108/6=-18

Answer 3

3p^2 + 50p - 72 = 0

 

You could solve by factoring:  (3p-4)(p+18)

 

p = 4/3

p = -18

 

Or you can use the quadratic formula, where a = 3, p = 50, and c = -72

 

(-b +/- square root of (b^2 - 4ac)) / 2a

 

(-50 +/- square root of (2500 + 864) / 6

(-50 +/- 58) / 6

Roots are 8/6 = 4/3 and -108/6 = -18