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}

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} : 4

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

The solutions to the quadratic equation are:

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

∣ 16 x + 20 ∣ = 8

e^{l n (5)}

3 a^{2} - 10 a - 8

x - 1 \leq 6

\frac{6}{( - \frac{4}{3} )}

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