What are the solutions to the equation -2w^2-12w-9=0 ?

The solutions to the equation -2w^2-12w-9=0 are w=-(3(2+sqrt(2)))/2 ,w=-(3(2-sqrt(2)))/2

Find the zeros of -2w^2-12w-9=0

The zeros of -2w^2-12w-9=0 are w=-(3(2+sqrt(2)))/2 ,w=-(3(2-sqrt(2)))/2

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- 2 w^{2} - 12 w - 9 = 0

w = - \frac{3 ( 2 + 2 )}{2}, w = - \frac{3 ( 2 - 2 )}{2}

Decimal Notation

w = - 5.12132 \dots, w = - 0.87867 \dots

Solution steps

- 2 w^{2} - 12 w - 9 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 ( - 2 ) ( - 9 )}{2 ( - 2 )}

(- 12)^{2} - 4 (- 2) (- 9) = 62

w_{1, 2} = \frac{- ( - 12 ) \pm 6 2}{2 ( - 2 )}

Separate the solutions

w_{1} = \frac{- ( - 12 ) + 6 2}{2 ( - 2 )}, w_{2} = \frac{- ( - 12 ) - 6 2}{2 ( - 2 )}

w = \frac{- ( - 12 ) + 6 2}{2 ( - 2 )} : - \frac{3 ( 2 + 2 )}{2}

w = \frac{- ( - 12 ) - 6 2}{2 ( - 2 )} : - \frac{3 ( 2 - 2 )}{2}

The solutions to the quadratic equation are:

w = - \frac{3 ( 2 + 2 )}{2}, w = - \frac{3 ( 2 - 2 )}{2}

4 n^{2} - 22 = 27

20 r^{2} + 37 r + 15

16 y^{2} + 8 y - 15

(m - 3) (m - 4) = 20

3 (2 + r) > 15 - 2 r

What are the solutions to the equation -2w^2-12w-9=0 ?
The solutions to the equation -2w^2-12w-9=0 are w=-(3(2+sqrt(2)))/2 ,w=-(3(2-sqrt(2)))/2
Find the zeros of -2w^2-12w-9=0
The zeros of -2w^2-12w-9=0 are w=-(3(2+sqrt(2)))/2 ,w=-(3(2-sqrt(2)))/2