What are the solutions to the equation 78x^2+47x-41=0 ?

The solutions to the equation 78x^2+47x-41=0 are x=(-47+sqrt(15001))/(156),x=(-47-sqrt(15001))/(156)

Find the zeros of 78x^2+47x-41=0

The zeros of 78x^2+47x-41=0 are x=(-47+sqrt(15001))/(156),x=(-47-sqrt(15001))/(156)

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78 x^{2} + 47 x - 41 = 0

x = \frac{- 47 + 15001}{156}, x = \frac{- 47 - 15001}{156}

Decimal Notation

x = 0.48383 \dots, x = - 1.08640 \dots

Solution steps

78 x^{2} + 47 x - 41 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 47 \pm 4 7 ^{2} - 4 \cdot 78 ( - 41 )}{2 \cdot 78}

4 7^{2} - 4 \cdot 78 (- 41) = 15001

x_{1, 2} = \frac{- 47 \pm 15001}{2 \cdot 78}

Separate the solutions

x_{1} = \frac{- 47 + 15001}{2 \cdot 78}, x_{2} = \frac{- 47 - 15001}{2 \cdot 78}

x = \frac{- 47 + 15001}{2 \cdot 78} : \frac{- 47 + 15001}{156}

x = \frac{- 47 - 15001}{2 \cdot 78} : \frac{- 47 - 15001}{156}

The solutions to the quadratic equation are:

x = \frac{- 47 + 15001}{156}, x = \frac{- 47 - 15001}{156}

9 x^{2} - 6 x - 3 x + 2

- 2 x + 3 \leq 5

3 \cdot 7

∣ 3 x - 2 ∣ > 16

(3 + 6)^{2}

What are the solutions to the equation 78x^2+47x-41=0 ?
The solutions to the equation 78x^2+47x-41=0 are x=(-47+sqrt(15001))/(156),x=(-47-sqrt(15001))/(156)
Find the zeros of 78x^2+47x-41=0
The zeros of 78x^2+47x-41=0 are x=(-47+sqrt(15001))/(156),x=(-47-sqrt(15001))/(156)