What are the solutions to the equation -w^2-3w-2=-5w^2 ?

The solutions to the equation -w^2-3w-2=-5w^2 are w=(3+sqrt(41))/8 ,w=(3-sqrt(41))/8

Find the zeros of -w^2-3w-2=-5w^2

The zeros of -w^2-3w-2=-5w^2 are w=(3+sqrt(41))/8 ,w=(3-sqrt(41))/8

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- w^{2} - 3 w - 2 = - 5 w^{2}

w = \frac{3 + 41}{8}, w = \frac{3 - 41}{8}

Decimal Notation

w = 1.17539 \dots, w = - 0.42539 \dots

Solution steps

- w^{2} - 3 w - 2 = - 5 w^{2}

Move

5 w^{2}

to the left side

4 w^{2} - 3 w - 2 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 4 ( - 2 )}{2 \cdot 4}

(- 3)^{2} - 4 \cdot 4 (- 2) = 41

w_{1, 2} = \frac{- ( - 3 ) \pm 41}{2 \cdot 4}

Separate the solutions

w_{1} = \frac{- ( - 3 ) + 41}{2 \cdot 4}, w_{2} = \frac{- ( - 3 ) - 41}{2 \cdot 4}

w = \frac{- ( - 3 ) + 41}{2 \cdot 4} : \frac{3 + 41}{8}

w = \frac{- ( - 3 ) - 41}{2 \cdot 4} : \frac{3 - 41}{8}

The solutions to the quadratic equation are:

w = \frac{3 + 41}{8}, w = \frac{3 - 41}{8}

n^{2} - 6 n - 7 = 0

x^{2} - 12 x + 64 = 0

b^{2} + 3 b = 0

x^{2} + 6 x - 520 = 0

x^{2} + 6 x + 8 = 3

What are the solutions to the equation -w^2-3w-2=-5w^2 ?
The solutions to the equation -w^2-3w-2=-5w^2 are w=(3+sqrt(41))/8 ,w=(3-sqrt(41))/8
Find the zeros of -w^2-3w-2=-5w^2
The zeros of -w^2-3w-2=-5w^2 are w=(3+sqrt(41))/8 ,w=(3-sqrt(41))/8