Second Degree Equations (2 Problems)

Hi, so I've got these 2 questions that I couldnt figure out!
1:
find C which makes 3X^2+12X+C=0 :
a)Two similar Roots (Delta = 0)
b)Two Different Roots (Delta > 0)
c)Two Imaginary Roots (Delta < 0 )

2:h=3t^2+25t :
Find t when h=18


and could you remind me of the formula for finding X using the Delta? It was something like ab+sqrt Delta / 2ab or something like that! thanks in advance!

10 Points For Best ANSWER!

1.
Are you familiar with the quadratic formula?

Applying that we get the roots

r = (-12 ± √(144 - 12C)) / 6

What concerns us is the number under the radical, 144 - 12C.

If 144 = 12C ⇔ C = 12, then the roots will be identical.
If 144 < 12C ⇔ C > 12, then the roots will be imaginary (since we have a negative under a square root).
And the remaining possibility is that C < 12, which will give us different roots.

#2.
h = 3t^2 + 25t
18 = 3t^2 + 25t

By the quadratic formula,
t = (-25 ± √(625 - 4(3)(-18)) ) / 6
t = (-25 ± 29) / 6
t = (4/6) or t = (-64/6)

Check my arithmetic to be sure!

Okay...I am not sure about the first one as our education systems are not the same but I can show you how to go about the second one
replace h with 18 and get
18=3t^2+25t
re-arrange the equation to get
3t^2+25t-18=0
use either the quadratic formula or completing the square method or just factor out to get the answer
I factored out and got
3t^2+27t-2t-18=0
3t(t+9)-2(t+9)=0
(3t-2)(t+9)=0
(t+9)=0
t=-9
(3t-2)=0
3t=2
t=0.6667
therefore t=-9 or 0.6667
hope you have understood...sorry to disappoint on the first one :-(

Each of these is a quadratic equation. Place in standard quadratic equation form and use the quadratic formula.

For the standard quadratic equation ax² + bx + c = 0, the solutions are

[-b + √(b² ‒ 4ac)]/(2a) or x = [-b ‒ √(b² ‒ 4ac)]/(2a).

The number under the radical, b² ‒ 4ac is the “discriminant.”