Question

8.Find the value of c that makes the expression be a perfect square trinomial p^2-30p+c

Answer 1

Hope this is the right one.

Problem 8
p^2 - 30p + c
Step One
Take 1/2 of - 30
1/2 * -30 = - 15

Step 2
Square -15
(-15)^2 = 225
c = 225

Problem Nine
$\text{x = }\dfrac{ -b \pm \sqrt{b^{2} - 4ac } }{2a}$
a = 1
b = 4
c = -15
$\text{x = }\dfrac{ -4 \pm \sqrt{4^{2} - 4*1*(-15) } }{2*1}$
$\text{x = }\dfrac{ -4 \pm \sqrt{\text{16} + \text{60} } }{2}$

x = [-4 +/- sqrt(76)] / 2
x = [-4 +/- 2*sqrt19]/2
x = [-4/2 +/- 2/2 sqrt[19]
x = - 2 +/- sqrt(19) 

x1 = - 2 + sqrt(19)
x2 = -2 - sqrt(19)

These two can be broken down more by finding the square root. I will leave them the way they are.  It's just a calculator question if you want it to go into decimal form.

Problem Ten

a = 1
b = 4
c = -32

The discriminate is sqrt(b^2 - 4ac)
D = sqrt(b^2 - 4ac)
D = sqrt(4^2 - 4(1)(-32)
D = sqrt(16 - - 128)
D = sqrt(16 + 128)
D = sqrt(144)
D = +/- 12

Since D can equal + or minus 12 there must be 2 possible (and different) roots. As a matter of fact, this quadratic can be factored.
(x + 8)(x - 4) = y 
But that' s not what you were asked for.
The discriminate is >  0 so the roots are going to be real.
Answer; The discriminate is > 0 so there will be 2 real different roots.