What are the solutions to the equation 2x^2+7x=-1 ?

The solutions to the equation 2x^2+7x=-1 are x=(-7+sqrt(41))/4 ,x=(-7-sqrt(41))/4

Find the zeros of 2x^2+7x=-1

The zeros of 2x^2+7x=-1 are x=(-7+sqrt(41))/4 ,x=(-7-sqrt(41))/4

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2 x^{2} + 7 x = - 1

x = \frac{- 7 + 41}{4}, x = \frac{- 7 - 41}{4}

Decimal Notation

x = - 0.14921 \dots, x = - 3.35078 \dots

Solution steps

2 x^{2} + 7 x = - 1

Move

1

to the left side

2 x^{2} + 7 x + 1 = 0

Solve with the quadratic formula

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

7^{2} - 4 \cdot 2 \cdot 1 = 41

x_{1, 2} = \frac{- 7 \pm 41}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 7 + 41}{2 \cdot 2}, x_{2} = \frac{- 7 - 41}{2 \cdot 2}

x = \frac{- 7 + 41}{2 \cdot 2} : \frac{- 7 + 41}{4}

x = \frac{- 7 - 41}{2 \cdot 2} : \frac{- 7 - 41}{4}

The solutions to the quadratic equation are:

x = \frac{- 7 + 41}{4}, x = \frac{- 7 - 41}{4}

4 x^{2} - 40 x + 84 = 0

x^{2} + 4 x + (- 1) = 0

x, f = x^{2} + y^{2} + x (y - z) - 4 x - 6 y + 4 z + 30

m^{2} - 4 m - 32 = 0

A, (A + X)^{2} = BC

What are the solutions to the equation 2x^2+7x=-1 ?
The solutions to the equation 2x^2+7x=-1 are x=(-7+sqrt(41))/4 ,x=(-7-sqrt(41))/4
Find the zeros of 2x^2+7x=-1
The zeros of 2x^2+7x=-1 are x=(-7+sqrt(41))/4 ,x=(-7-sqrt(41))/4