What are the solutions to the equation 12x^2+5x-4=0 ?

The solutions to the equation 12x^2+5x-4=0 are x=(-5+sqrt(217))/(24),x=(-5-sqrt(217))/(24)

Find the zeros of 12x^2+5x-4=0

The zeros of 12x^2+5x-4=0 are x=(-5+sqrt(217))/(24),x=(-5-sqrt(217))/(24)

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12 x^{2} + 5 x - 4 = 0

x = \frac{- 5 + 217}{24}, x = \frac{- 5 - 217}{24}

Decimal Notation

x = 0.40545 \dots, x = - 0.82212 \dots

Solution steps

12 x^{2} + 5 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 12 ( - 4 )}{2 \cdot 12}

5^{2} - 4 \cdot 12 (- 4) = 217

x_{1, 2} = \frac{- 5 \pm 217}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- 5 + 217}{2 \cdot 12}, x_{2} = \frac{- 5 - 217}{2 \cdot 12}

x = \frac{- 5 + 217}{2 \cdot 12} : \frac{- 5 + 217}{24}

x = \frac{- 5 - 217}{2 \cdot 12} : \frac{- 5 - 217}{24}

The solutions to the quadratic equation are:

x = \frac{- 5 + 217}{24}, x = \frac{- 5 - 217}{24}

0 = - x^{2} + 100 x

4 x^{2} - 9 = 2 x^{2} + 41

- 4 x^{2} + 16 x + 5 = 0

5 u^{2} + u = 4

z^{2} + 6 z = r^{2} - 9

What are the solutions to the equation 12x^2+5x-4=0 ?
The solutions to the equation 12x^2+5x-4=0 are x=(-5+sqrt(217))/(24),x=(-5-sqrt(217))/(24)
Find the zeros of 12x^2+5x-4=0
The zeros of 12x^2+5x-4=0 are x=(-5+sqrt(217))/(24),x=(-5-sqrt(217))/(24)