What are the solutions to the equation -11a^2+32a+256=0 ?

The solutions to the equation -11a^2+32a+256=0 are a=-(16(2sqrt(3)-1))/(11),a=(16(1+2sqrt(3)))/(11)

Find the zeros of -11a^2+32a+256=0

The zeros of -11a^2+32a+256=0 are a=-(16(2sqrt(3)-1))/(11),a=(16(1+2sqrt(3)))/(11)

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- 11 a^{2} + 32 a + 256 = 0

a = - \frac{16 ( 2 3 - 1 )}{11}, a = \frac{16 ( 1 + 2 3 )}{11}

Decimal Notation

a = - 3.58414 \dots, a = 6.49323 \dots

Solution steps

- 11 a^{2} + 32 a + 256 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 ( - 11 ) \cdot 256}{2 ( - 11 )}

3 2^{2} - 4 (- 11) \cdot 256 = 643

a_{1, 2} = \frac{- 32 \pm 64 3}{2 ( - 11 )}

Separate the solutions

a_{1} = \frac{- 32 + 64 3}{2 ( - 11 )}, a_{2} = \frac{- 32 - 64 3}{2 ( - 11 )}

a = \frac{- 32 + 64 3}{2 ( - 11 )} : - \frac{16 ( 2 3 - 1 )}{11}

a = \frac{- 32 - 64 3}{2 ( - 11 )} : \frac{16 ( 1 + 2 3 )}{11}

The solutions to the quadratic equation are:

a = - \frac{16 ( 2 3 - 1 )}{11}, a = \frac{16 ( 1 + 2 3 )}{11}

- 7 x^{2} - 21 x - 5

3 x + 5 = 4 x^{2}

t, 2 a t^{2} - 6 a t - a + 2 t b - 3 b + 2 c = t^{2} + 3

2 \cdot ((0.145 - 2 x)^{2}) = (0.0262 + x) \cdot (0.074 + x)

\frac{x ^{2}}{16} - \frac{4}{49} = 0

What are the solutions to the equation -11a^2+32a+256=0 ?
The solutions to the equation -11a^2+32a+256=0 are a=-(16(2sqrt(3)-1))/(11),a=(16(1+2sqrt(3)))/(11)
Find the zeros of -11a^2+32a+256=0
The zeros of -11a^2+32a+256=0 are a=-(16(2sqrt(3)-1))/(11),a=(16(1+2sqrt(3)))/(11)