What are the solutions to the equation 2x^2+7x-49=0 ?

The solutions to the equation 2x^2+7x-49=0 are x= 7/2 ,x=-7

Find the zeros of 2x^2+7x-49=0

The zeros of 2x^2+7x-49=0 are x= 7/2 ,x=-7

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2 x^{2} + 7 x - 49 = 0

x = \frac{7}{2}, x = - 7

Decimal Notation

x = 3.5, x = - 7

Solution steps

2 x^{2} + 7 x - 49 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 ( - 49 )}{2 \cdot 2}

7^{2} - 4 \cdot 2 (- 49) = 21

x_{1, 2} = \frac{- 7 \pm 21}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 7 + 21}{2 \cdot 2}, x_{2} = \frac{- 7 - 21}{2 \cdot 2}

x = \frac{- 7 + 21}{2 \cdot 2} : \frac{7}{2}

x = \frac{- 7 - 21}{2 \cdot 2} : - 7

The solutions to the quadratic equation are:

x = \frac{7}{2}, x = - 7

a, a^{2} - 2 a = b^{2} - 2 b

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

x^{2} + 5 x + 5 + 8 = 0

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

4 x - x^{2} - 3 = 0

What are the solutions to the equation 2x^2+7x-49=0 ?
The solutions to the equation 2x^2+7x-49=0 are x= 7/2 ,x=-7
Find the zeros of 2x^2+7x-49=0
The zeros of 2x^2+7x-49=0 are x= 7/2 ,x=-7