What are the solutions to the equation (2w+5)w=42 ?

The solutions to the equation (2w+5)w=42 are w= 7/2 ,w=-6

Find the zeros of (2w+5)w=42

The zeros of (2w+5)w=42 are w= 7/2 ,w=-6

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(2 w + 5) w = 42

w = \frac{7}{2}, w = - 6

Decimal Notation

w = 3.5, w = - 6

Solution steps

(2 w + 5) w = 42

Expand

(2 w + 5) w : 2 w^{2} + 5 w

2 w^{2} + 5 w = 42

Move

42

to the left side

2 w^{2} + 5 w - 42 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 ( - 42 )}{2 \cdot 2}

5^{2} - 4 \cdot 2 (- 42) = 19

w_{1, 2} = \frac{- 5 \pm 19}{2 \cdot 2}

Separate the solutions

w_{1} = \frac{- 5 + 19}{2 \cdot 2}, w_{2} = \frac{- 5 - 19}{2 \cdot 2}

w = \frac{- 5 + 19}{2 \cdot 2} : \frac{7}{2}

w = \frac{- 5 - 19}{2 \cdot 2} : - 6

The solutions to the quadratic equation are:

w = \frac{7}{2}, w = - 6

\frac{2}{3} (16)^{\frac{3}{2}}

x^{2} + (\frac{1}{7})^{2} = 1

- 10 + 6

4 x^{2} - 5 = 3 x + 4

\frac{d}{4} > 6

What are the solutions to the equation (2w+5)w=42 ?
The solutions to the equation (2w+5)w=42 are w= 7/2 ,w=-6
Find the zeros of (2w+5)w=42
The zeros of (2w+5)w=42 are w= 7/2 ,w=-6