What are the solutions to the equation quadraticformula 2x^2+x-7=0 ?

The solutions to the equation quadraticformula 2x^2+x-7=0 are x=(-1+sqrt(57))/4 ,x=(-1-sqrt(57))/4

Find the zeros of quadraticformula 2x^2+x-7=0

The zeros of quadraticformula 2x^2+x-7=0 are x=(-1+sqrt(57))/4 ,x=(-1-sqrt(57))/4

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2 x^{2} + x - 7 = 0

x = \frac{- 1 + 57}{4}, x = \frac{- 1 - 57}{4}

Solution steps

2 x^{2} + x - 7 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 2 (- 7) = 57

x_{1, 2} = \frac{- 1 \pm 57}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 1 + 57}{2 \cdot 2}, x_{2} = \frac{- 1 - 57}{2 \cdot 2}

x = \frac{- 1 + 57}{2 \cdot 2} : \frac{- 1 + 57}{4}

x = \frac{- 1 - 57}{2 \cdot 2} : \frac{- 1 - 57}{4}

The solutions to the quadratic equation are:

x = \frac{- 1 + 57}{4}, x = \frac{- 1 - 57}{4}

x, 8 < x (7 - x)

∣ - 5.8 ∣

x^{2} - 2 x - 10 = 0

c, - 32 c + 12 \leq - 66 c - 16

31%

What are the solutions to the equation quadraticformula 2x^2+x-7=0 ?
The solutions to the equation quadraticformula 2x^2+x-7=0 are x=(-1+sqrt(57))/4 ,x=(-1-sqrt(57))/4
Find the zeros of quadraticformula 2x^2+x-7=0
The zeros of quadraticformula 2x^2+x-7=0 are x=(-1+sqrt(57))/4 ,x=(-1-sqrt(57))/4