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

The solutions to the equation 2x^2+x-5=0 are x=(-1+sqrt(41))/4 ,x=(-1-sqrt(41))/4

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

The zeros of 2x^2+x-5=0 are x=(-1+sqrt(41))/4 ,x=(-1-sqrt(41))/4

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

x = \frac{- 1 + 41}{4}, x = \frac{- 1 - 41}{4}

Decimal Notation

x = 1.35078 \dots, x = - 1.85078 \dots

Solution steps

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

Solve with the quadratic formula

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

1^{2} - 4 \cdot 2 (- 5) = 41

x_{1, 2} = \frac{- 1 \pm 41}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 1 + 41}{2 \cdot 2}, x_{2} = \frac{- 1 - 41}{2 \cdot 2}

x = \frac{- 1 + 41}{2 \cdot 2} : \frac{- 1 + 41}{4}

x = \frac{- 1 - 41}{2 \cdot 2} : \frac{- 1 - 41}{4}

The solutions to the quadratic equation are:

x = \frac{- 1 + 41}{4}, x = \frac{- 1 - 41}{4}

\frac{12 !}{( 12 - 2 ) ! \cdot 2 !}

\frac{- 1 + 5 i}{1 - i}

50 m^{2} - 5 m - 1 = 0

4 (x - 3) \frac{1}{2} = 8

\frac{1}{8 x ^{- 2} y ^{- 6}}

What are the solutions to the equation 2x^2+x-5=0 ?
The solutions to the equation 2x^2+x-5=0 are x=(-1+sqrt(41))/4 ,x=(-1-sqrt(41))/4
Find the zeros of 2x^2+x-5=0
The zeros of 2x^2+x-5=0 are x=(-1+sqrt(41))/4 ,x=(-1-sqrt(41))/4