Questions: Solve for (t). [ d=-16 t^2+15 t ] What is the answer? ( t=frac-15 pm sqrt225-64 d-32 )

Solution Steps

To solve for \( t \) in the equation \( d = -16t^2 + 15t \), we recognize it as a quadratic equation in the form \( at^2 + bt + c = 0 \). We can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the values of \( t \). Here, \( a = -16 \), \( b = 15 \), and \( c = -d \).

The given equation is \( d = -16t^2 + 15t \). This is a quadratic equation in the standard form \( at^2 + bt + c = 0 \), where \( a = -16 \), \( b = 15 \), and \( c = -d \).

To solve for \( t \), we use the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substituting the values of \( a \), \( b \), and \( c \), we have: \[ t = \frac{-15 \pm \sqrt{15^2 - 4(-16)(-d)}}{2(-16)} \]

The discriminant is given by: \[ b^2 - 4ac = 15^2 - 4(-16)(-d) = 225 - 64d \] For \( d = 0 \), the discriminant simplifies to: \[ 225 \]

Substitute the discriminant back into the quadratic formula: \[ t = \frac{-15 \pm \sqrt{225}}{-32} \] Calculate the two possible values for \( t \): \[ t_1 = \frac{-15 + 15}{-32} = 0 \] \[ t_2 = \frac{-15 - 15}{-32} = 0.9375 \]

Final Answer