What are the solutions to the equation 32x^2-12x-4=0 ?

The solutions to the equation 32x^2-12x-4=0 are x=(3+sqrt(41))/(16),x=(3-sqrt(41))/(16)

Find the zeros of 32x^2-12x-4=0

The zeros of 32x^2-12x-4=0 are x=(3+sqrt(41))/(16),x=(3-sqrt(41))/(16)

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32 x^{2} - 12 x - 4 = 0

x = \frac{3 + 41}{16}, x = \frac{3 - 41}{16}

Decimal Notation

x = 0.58769 \dots, x = - 0.21269 \dots

Solution steps

32 x^{2} - 12 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 32 ( - 4 )}{2 \cdot 32}

(- 12)^{2} - 4 \cdot 32 (- 4) = 441

x_{1, 2} = \frac{- ( - 12 ) \pm 4 41}{2 \cdot 32}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 4 41}{2 \cdot 32}, x_{2} = \frac{- ( - 12 ) - 4 41}{2 \cdot 32}

x = \frac{- ( - 12 ) + 4 41}{2 \cdot 32} : \frac{3 + 41}{16}

x = \frac{- ( - 12 ) - 4 41}{2 \cdot 32} : \frac{3 - 41}{16}

The solutions to the quadratic equation are:

x = \frac{3 + 41}{16}, x = \frac{3 - 41}{16}

- x^{2} = - 16

- x^{2} - 5 x - 2 = 0

- 2 x^{2} + 24 x - 72 = 0

x^{2} - 13 x - 22 = 0

9 x^{2} - 4 x - 7 = 0

What are the solutions to the equation 32x^2-12x-4=0 ?
The solutions to the equation 32x^2-12x-4=0 are x=(3+sqrt(41))/(16),x=(3-sqrt(41))/(16)
Find the zeros of 32x^2-12x-4=0
The zeros of 32x^2-12x-4=0 are x=(3+sqrt(41))/(16),x=(3-sqrt(41))/(16)