What do I have wrong in here?

A stone is thrown vertically upward at a speed of 60.0m/s.
a) How long after being thrown is it 175m above ground?

So i did -
d = V1t + 1/2at^2
175 = 60t + 1/2 (10)t^2 (rearranged and divided it all by 5)
0 = t^2 + 12t - 35
And I can't get the factors.

We were asked to consider acceleration as 10.0m/s^2

How do you know if a quadratic is a perfect square?

Hey, if velocity up is +, and distance up is +, then gravity down is -

ok so call g = 9.8 which you are calling 10 close enough DOWN

175 = 60 t - 5 t^2
t^2 - 12 t + 35 = 0
(t-5)(t-7) = 0
t = 5 and t = 7
5 seconds on the way up, 7 seconds on the way down again.

To find out if a quadratic equation is a perfect square, you can use the discriminant. In the quadratic equation ax^2 + bx + c = 0, the discriminant is represented as Δ = b^2 - 4ac.

In your case, the quadratic equation is t^2 + 12t - 35 = 0. Comparing it with the standard quadratic equation ax^2 + bx + c = 0, we have a = 1, b = 12, and c = -35.

Now, calculate the discriminant Δ by substituting the values into the formula: Δ = (12)^2 - 4(1)(-35).

Δ = 144 + 140

Δ = 284

If Δ is a perfect square, then the quadratic equation is a perfect square. However, in this case, 284 is not a perfect square. Therefore, the quadratic equation t^2 + 12t - 35 = 0 is not a perfect square.

To solve the equation, you can use the quadratic formula: t = (-b ± √Δ) / (2a).

Substituting the values into the quadratic formula, you will get:

t = (-12 ± √284) / (2(1))

Simplifying further:

t = (-12 ± √284) / 2

t = (-12 ± √(4 * 71)) / 2

t = (-12 ± 2√71) / 2

t = -6 ± √71

Therefore, the solutions to the equation t^2 + 12t - 35 = 0 are:

t = -6 + √71

and

t = -6 - √71