What are the solutions to the equation 4w^2+28w+49=4 ?

The solutions to the equation 4w^2+28w+49=4 are w=-5/2 ,w=-9/2

Find the zeros of 4w^2+28w+49=4

The zeros of 4w^2+28w+49=4 are w=-5/2 ,w=-9/2

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4 w^{2} + 28 w + 49 = 4

w = - \frac{5}{2}, w = - \frac{9}{2}

Decimal Notation

w = - 2.5, w = - 4.5

Solution steps

4 w^{2} + 28 w + 49 = 4

Move

4

to the left side

4 w^{2} + 28 w + 45 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 4 \cdot 45}{2 \cdot 4}

2 8^{2} - 4 \cdot 4 \cdot 45 = 8

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

Separate the solutions

w_{1} = \frac{- 28 + 8}{2 \cdot 4}, w_{2} = \frac{- 28 - 8}{2 \cdot 4}

w = \frac{- 28 + 8}{2 \cdot 4} : - \frac{5}{2}

w = \frac{- 28 - 8}{2 \cdot 4} : - \frac{9}{2}

The solutions to the quadratic equation are:

w = - \frac{5}{2}, w = - \frac{9}{2}

49 p^{2} - 121 q^{2}

∣ x - 3 ∣ < 6

\frac{1}{\frac{1}{6} + \frac{1}{12}}

\frac{4 + i}{- 5 - 2 i}

- 4 - 4 i

What are the solutions to the equation 4w^2+28w+49=4 ?
The solutions to the equation 4w^2+28w+49=4 are w=-5/2 ,w=-9/2
Find the zeros of 4w^2+28w+49=4
The zeros of 4w^2+28w+49=4 are w=-5/2 ,w=-9/2