What are the solutions to the equation 4w^2+4w+1=49 ?

The solutions to the equation 4w^2+4w+1=49 are w=3,w=-4

Find the zeros of 4w^2+4w+1=49

The zeros of 4w^2+4w+1=49 are w=3,w=-4

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4 w^{2} + 4 w + 1 = 49

w = 3, w = - 4

Solution steps

4 w^{2} + 4 w + 1 = 49

Move

49

to the left side

4 w^{2} + 4 w - 48 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 4 ( - 48 )}{2 \cdot 4}

4^{2} - 4 \cdot 4 (- 48) = 28

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

Separate the solutions

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

w = \frac{- 4 + 28}{2 \cdot 4} : 3

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

The solutions to the quadratic equation are:

w = 3, w = - 4

2 z^{2} + z - 7 = 0

\frac{x + 2}{8} = \frac{3}{4}

3 x (x + 2) = 0

n^{2} + 15 n + 56

3 x - 2 = x + 8

What are the solutions to the equation 4w^2+4w+1=49 ?
The solutions to the equation 4w^2+4w+1=49 are w=3,w=-4
Find the zeros of 4w^2+4w+1=49
The zeros of 4w^2+4w+1=49 are w=3,w=-4