What are the solutions to the equation 2w^2+22w=24 ?

The solutions to the equation 2w^2+22w=24 are w=1,w=-12

Find the zeros of 2w^2+22w=24

The zeros of 2w^2+22w=24 are w=1,w=-12

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2 w^{2} + 22 w = 24

w = 1, w = - 12

Solution steps

2 w^{2} + 22 w = 24

Move

24

to the left side

2 w^{2} + 22 w - 24 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 22 \pm 2 2 ^{2} - 4 \cdot 2 ( - 24 )}{2 \cdot 2}

2 2^{2} - 4 \cdot 2 (- 24) = 26

w_{1, 2} = \frac{- 22 \pm 26}{2 \cdot 2}

Separate the solutions

w_{1} = \frac{- 22 + 26}{2 \cdot 2}, w_{2} = \frac{- 22 - 26}{2 \cdot 2}

w = \frac{- 22 + 26}{2 \cdot 2} : 1

w = \frac{- 22 - 26}{2 \cdot 2} : - 12

The solutions to the quadratic equation are:

w = 1, w = - 12

(k + 1)^{2} = 9

2 x^{2} - 8 x - 11 = x^{2} - 5 x - 7

2 x^{2} = 6 x + 5

x^{2} + 18 x = - 68

9 x^{2} - 24 x = 0

What are the solutions to the equation 2w^2+22w=24 ?
The solutions to the equation 2w^2+22w=24 are w=1,w=-12
Find the zeros of 2w^2+22w=24
The zeros of 2w^2+22w=24 are w=1,w=-12