What are the solutions to the equation 4w^2+12w=-5 ?

The solutions to the equation 4w^2+12w=-5 are w=-1/2 ,w=-5/2

Find the zeros of 4w^2+12w=-5

The zeros of 4w^2+12w=-5 are w=-1/2 ,w=-5/2

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4 w^{2} + 12 w = - 5

w = - \frac{1}{2}, w = - \frac{5}{2}

Decimal Notation

w = - 0.5, w = - 2.5

Solution steps

4 w^{2} + 12 w = - 5

Move

5

to the left side

4 w^{2} + 12 w + 5 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 4 \cdot 5}{2 \cdot 4}

1 2^{2} - 4 \cdot 4 \cdot 5 = 8

w_{1, 2} = \frac{- 12 \pm 8}{2 \cdot 4}

Separate the solutions

w_{1} = \frac{- 12 + 8}{2 \cdot 4}, w_{2} = \frac{- 12 - 8}{2 \cdot 4}

w = \frac{- 12 + 8}{2 \cdot 4} : - \frac{1}{2}

w = \frac{- 12 - 8}{2 \cdot 4} : - \frac{5}{2}

The solutions to the quadratic equation are:

w = - \frac{1}{2}, w = - \frac{5}{2}

5 - 2 x < 7

13 - \frac{3}{2} x = 37

x^{3} + 5 x

\frac{3}{4} (2)^{\frac{4}{3}}

- 4 (m + 6) = - 8

What are the solutions to the equation 4w^2+12w=-5 ?
The solutions to the equation 4w^2+12w=-5 are w=-1/2 ,w=-5/2
Find the zeros of 4w^2+12w=-5
The zeros of 4w^2+12w=-5 are w=-1/2 ,w=-5/2