What are the solutions to the equation 151=235t-16t^2 ?

The solutions to the equation 151=235t-16t^2 are t=(235-sqrt(45561))/(32),t=(235+sqrt(45561))/(32)

Find the zeros of 151=235t-16t^2

The zeros of 151=235t-16t^2 are t=(235-sqrt(45561))/(32),t=(235+sqrt(45561))/(32)

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151 = 235 t - 16 t^{2}

t = \frac{235 - 45561}{32}, t = \frac{235 + 45561}{32}

Decimal Notation

t = 0.67343 \dots, t = 14.01406 \dots

Solution steps

151 = 235 t - 16 t^{2}

Switch sides

235 t - 16 t^{2} = 151

Move

151

to the left side

235 t - 16 t^{2} - 151 = 0

Write in the standard form

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

- 16 t^{2} + 235 t - 151 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 235 \pm 23 5 ^{2} - 4 ( - 16 ) ( - 151 )}{2 ( - 16 )}

23 5^{2} - 4 (- 16) (- 151) = 45561

t_{1, 2} = \frac{- 235 \pm 45561}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 235 + 45561}{2 ( - 16 )}, t_{2} = \frac{- 235 - 45561}{2 ( - 16 )}

t = \frac{- 235 + 45561}{2 ( - 16 )} : \frac{235 - 45561}{32}

t = \frac{- 235 - 45561}{2 ( - 16 )} : \frac{235 + 45561}{32}

The solutions to the quadratic equation are:

t = \frac{235 - 45561}{32}, t = \frac{235 + 45561}{32}

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567 m^{2} - 343 = 0

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What are the solutions to the equation 151=235t-16t^2 ?
The solutions to the equation 151=235t-16t^2 are t=(235-sqrt(45561))/(32),t=(235+sqrt(45561))/(32)
Find the zeros of 151=235t-16t^2
The zeros of 151=235t-16t^2 are t=(235-sqrt(45561))/(32),t=(235+sqrt(45561))/(32)