What are the solutions to the equation 15w^2+28w+3=-2 ?

The solutions to the equation 15w^2+28w+3=-2 are w=-1/5 ,w=-5/3

Find the zeros of 15w^2+28w+3=-2

The zeros of 15w^2+28w+3=-2 are w=-1/5 ,w=-5/3

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15 w^{2} + 28 w + 3 = - 2

w = - \frac{1}{5}, w = - \frac{5}{3}

Solution steps

15 w^{2} + 28 w + 3 = - 2

Move

2

to the left side

15 w^{2} + 28 w + 5 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 15 \cdot 5}{2 \cdot 15}

2 8^{2} - 4 \cdot 15 \cdot 5 = 22

w_{1, 2} = \frac{- 28 \pm 22}{2 \cdot 15}

Separate the solutions

w_{1} = \frac{- 28 + 22}{2 \cdot 15}, w_{2} = \frac{- 28 - 22}{2 \cdot 15}

w = \frac{- 28 + 22}{2 \cdot 15} : - \frac{1}{5}

w = \frac{- 28 - 22}{2 \cdot 15} : - \frac{5}{3}

The solutions to the quadratic equation are:

w = - \frac{1}{5}, w = - \frac{5}{3}

\frac{x + y}{x - y}

6 x - 8 = 4 x - 5

42 = 7 r

x^{2} - 10 x y - 11 y^{2}

16 x + 4

What are the solutions to the equation 15w^2+28w+3=-2 ?
The solutions to the equation 15w^2+28w+3=-2 are w=-1/5 ,w=-5/3
Find the zeros of 15w^2+28w+3=-2
The zeros of 15w^2+28w+3=-2 are w=-1/5 ,w=-5/3