What are the solutions to the equation 2w^2-w=28 ?

The solutions to the equation 2w^2-w=28 are w=4,w=-7/2

Find the zeros of 2w^2-w=28

The zeros of 2w^2-w=28 are w=4,w=-7/2

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2 w^{2} - w = 28

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

Solution steps

2 w^{2} - w = 28

Move

28

to the left side

2 w^{2} - w - 28 = 0

Solve with the quadratic formula

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

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

w_{1, 2} = \frac{- ( - 1 ) \pm 15}{2 \cdot 2}

Separate the solutions

w_{1} = \frac{- ( - 1 ) + 15}{2 \cdot 2}, w_{2} = \frac{- ( - 1 ) - 15}{2 \cdot 2}

w = \frac{- ( - 1 ) + 15}{2 \cdot 2} : 4

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

The solutions to the quadratic equation are:

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

∣ x - 6 ∣ \leq 6

\frac{24 x ^{2}}{5 x y}

\frac{( x - 4 )}{( x - 1 )} > \frac{x}{3} + 4

(x - 4) (x + 6) < 0

2 (- \frac{4}{3})^{2} + 5 (- \frac{4}{3}) + 6

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