What are the solutions to the equation 97=215t-16t^2 ?

The solutions to the equation 97=215t-16t^2 are t=(215-sqrt(40017))/(32),t=(215+sqrt(40017))/(32)

Find the zeros of 97=215t-16t^2

The zeros of 97=215t-16t^2 are t=(215-sqrt(40017))/(32),t=(215+sqrt(40017))/(32)

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97 = 215 t - 16 t^{2}

t = \frac{215 - 40017}{32}, t = \frac{215 + 40017}{32}

Decimal Notation

t = 0.46742 \dots, t = 12.97007 \dots

Solution steps

97 = 215 t - 16 t^{2}

Switch sides

215 t - 16 t^{2} = 97

Move

97

to the left side

215 t - 16 t^{2} - 97 = 0

Write in the standard form

a x^{2} + b x + c = 0

- 16 t^{2} + 215 t - 97 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 215 \pm 21 5 ^{2} - 4 ( - 16 ) ( - 97 )}{2 ( - 16 )}

21 5^{2} - 4 (- 16) (- 97) = 40017

t_{1, 2} = \frac{- 215 \pm 40017}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 215 + 40017}{2 ( - 16 )}, t_{2} = \frac{- 215 - 40017}{2 ( - 16 )}

t = \frac{- 215 + 40017}{2 ( - 16 )} : \frac{215 - 40017}{32}

t = \frac{- 215 - 40017}{2 ( - 16 )} : \frac{215 + 40017}{32}

The solutions to the quadratic equation are:

t = \frac{215 - 40017}{32}, t = \frac{215 + 40017}{32}

3 x^{2} - 14 x + 8 > 0

- 4 a^{2} b - 5 a^{2} b

2 s^{2} + 3 s - 2

\frac{a b ^{2}}{a c} + \frac{c ^{2} b}{ab}

x^{2} + 80 = 0

What are the solutions to the equation 97=215t-16t^2 ?
The solutions to the equation 97=215t-16t^2 are t=(215-sqrt(40017))/(32),t=(215+sqrt(40017))/(32)
Find the zeros of 97=215t-16t^2
The zeros of 97=215t-16t^2 are t=(215-sqrt(40017))/(32),t=(215+sqrt(40017))/(32)