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

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

Find the zeros of quadraticformula 2x^2+x=5

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

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2 x^{2} + x = 5

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

Move

5

to the left side

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}

5 x^{2} + 400 = 0

(4 x + 9)^{2} = 36

x, y^{3} - 5 y = 4 x - x^{2} + 9

x^{2} + 5 x = 104

- 5 x^{2} + 80 x - 320 = 0

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