What are the solutions to the equation 2x^2+x-28=0 ?

The solutions to the equation 2x^2+x-28=0 are x= 7/2 ,x=-4

Find the zeros of 2x^2+x-28=0

The zeros of 2x^2+x-28=0 are x= 7/2 ,x=-4

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2 x^{2} + x - 28 = 0

x = \frac{7}{2}, x = - 4

Solution steps

2 x^{2} + x - 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 28 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 28) = 15

x_{1, 2} = \frac{- 1 \pm 15}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 1 + 15}{2 \cdot 2}, x_{2} = \frac{- 1 - 15}{2 \cdot 2}

x = \frac{- 1 + 15}{2 \cdot 2} : \frac{7}{2}

x = \frac{- 1 - 15}{2 \cdot 2} : - 4

The solutions to the quadratic equation are:

x = \frac{7}{2}, x = - 4

\frac{5}{9} (122 - 32)

4 - 20

25 x^{2} - 1 = 0

v + 10 < 15

4^{2} + x^{2} = 7^{2}

What are the solutions to the equation 2x^2+x-28=0 ?
The solutions to the equation 2x^2+x-28=0 are x= 7/2 ,x=-4
Find the zeros of 2x^2+x-28=0
The zeros of 2x^2+x-28=0 are x= 7/2 ,x=-4