What are the solutions to the equation 2x^{(2)}-x-28=0 ?

The solutions to the equation 2x^{(2)}-x-28=0 are x=4,x=-7/2

Find the zeros of 2x^{(2)}-x-28=0

The zeros of 2x^{(2)}-x-28=0 are x=4,x=-7/2

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2 x^{(2)} - x - 28 = 0

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

Decimal

x = 4, x = - 3.5

Solution steps

2 x^{(2)} - x - 28 = 0

Refine

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} : 4

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

The solutions to the quadratic equation are:

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

c (x) = 1500 + 5 x + 0.01 x^{2}

x^{2} + c x + b = 0

x, lo g_{10} (5) x^{2} - lo g_{10} (5 y) = 1

32 y^{2} + 32 y + 8 = 0

x^{2} = \frac{lo g _{10} ( 25 )}{lo g _{10} ( 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=4,x=-7/2
Find the zeros of 2x^{(2)}-x-28=0
The zeros of 2x^{(2)}-x-28=0 are x=4,x=-7/2