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0=-18+35t-16t^2 please solve

User Owens
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1 Answer

4 votes

Answer: t₁ = (-35 + √73) / (-32)

t₂ = (-35 - √73) / (-32)

Explanation:

To solve the quadratic equation:

0 = -18 + 35t - 16t^2

You can rearrange it into standard quadratic form, which is:

0 = -16t^2 + 35t - 18

Now, you can attempt to factor it or use the quadratic formula to find the values of t that satisfy this equation. Let's use the quadratic formula:

The quadratic formula is given by:

t = (-b ± √(b² - 4ac)) / (2a)

In this equation, a = -16, b = 35, and c = -18. Plug these values into the formula:

t = (-(35) ± √((35)² - 4(-16)(-18))) / (2(-16))

Now, calculate:

t = (-35 ± √(1225 - 1152)) / (-32)

t = (-35 ± √73) / (-32)

So, the solutions for t are:

t₁ = (-35 + √73) / (-32)

t₂ = (-35 - √73) / (-32)

These are the two values of t that satisfy the equation.

User Jure
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