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

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

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

The zeros of 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}

Decimal Notation

x = 1.63745 \dots, x = - 2.13745 \dots

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}

\frac{2 x + 3}{x - 3} > 1

∣ 6 x - 5 ∣ = 4

2 (x + y)^{2} - 9 (x + y) - 5

f^{2} - 3 f - 4

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