What are the solutions to the equation 2x^2+7x-51=-2 ?

The solutions to the equation 2x^2+7x-51=-2 are x= 7/2 ,x=-7

Find the zeros of 2x^2+7x-51=-2

The zeros of 2x^2+7x-51=-2 are x= 7/2 ,x=-7

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2 x^{2} + 7 x - 51 = - 2

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

Decimal Notation

x = 3.5, x = - 7

Solution steps

2 x^{2} + 7 x - 51 = - 2

Move

2

to the left side

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

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

x^{2} - 4 x + 70

B, (A \cap B) \cup (A \cap B) = A

192 - 112 x + 12 x^{2} = 0

64 - \frac{x ^{2}}{25} = 0

What are the solutions to the equation 2x^2+7x-51=-2 ?
The solutions to the equation 2x^2+7x-51=-2 are x= 7/2 ,x=-7
Find the zeros of 2x^2+7x-51=-2
The zeros of 2x^2+7x-51=-2 are x= 7/2 ,x=-7