What are the solutions to the equation -16t^2+288t+216=0 ?

The solutions to the equation -16t^2+288t+216=0 are t=-(3(sqrt(42)-6))/2 ,t=(3(6+sqrt(42)))/2

Find the zeros of -16t^2+288t+216=0

The zeros of -16t^2+288t+216=0 are t=-(3(sqrt(42)-6))/2 ,t=(3(6+sqrt(42)))/2

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- 16 t^{2} + 288 t + 216 = 0

t = - \frac{3 ( 42 - 6 )}{2}, t = \frac{3 ( 6 + 42 )}{2}

Decimal Notation

t = - 0.72111 \dots, t = 18.72111 \dots

Solution steps

- 16 t^{2} + 288 t + 216 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 288 \pm 28 8 ^{2} - 4 ( - 16 ) \cdot 216}{2 ( - 16 )}

28 8^{2} - 4 (- 16) \cdot 216 = 4842

t_{1, 2} = \frac{- 288 \pm 48 42}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 288 + 48 42}{2 ( - 16 )}, t_{2} = \frac{- 288 - 48 42}{2 ( - 16 )}

t = \frac{- 288 + 48 42}{2 ( - 16 )} : - \frac{3 ( 42 - 6 )}{2}

t = \frac{- 288 - 48 42}{2 ( - 16 )} : \frac{3 ( 6 + 42 )}{2}

The solutions to the quadratic equation are:

t = - \frac{3 ( 42 - 6 )}{2}, t = \frac{3 ( 6 + 42 )}{2}

a^{2} + 13 a + 40 = 0

(x + 2) (x - 1) = 4

4 k (k + 2) = 2

(- x^{2} + 2 x + 1) = - 1 (x - 2 x - 1)

x (x - 4) = - 5

What are the solutions to the equation -16t^2+288t+216=0 ?
The solutions to the equation -16t^2+288t+216=0 are t=-(3(sqrt(42)-6))/2 ,t=(3(6+sqrt(42)))/2
Find the zeros of -16t^2+288t+216=0
The zeros of -16t^2+288t+216=0 are t=-(3(sqrt(42)-6))/2 ,t=(3(6+sqrt(42)))/2