What are the solutions to the equation 4w^2+4w+1=169 ?

The solutions to the equation 4w^2+4w+1=169 are w=6,w=-7

Find the zeros of 4w^2+4w+1=169

The zeros of 4w^2+4w+1=169 are w=6,w=-7

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4 w^{2} + 4 w + 1 = 169

w = 6, w = - 7

Solution steps

4 w^{2} + 4 w + 1 = 169

Move

169

to the left side

4 w^{2} + 4 w - 168 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 4 ( - 168 )}{2 \cdot 4}

4^{2} - 4 \cdot 4 (- 168) = 52

w_{1, 2} = \frac{- 4 \pm 52}{2 \cdot 4}

Separate the solutions

w_{1} = \frac{- 4 + 52}{2 \cdot 4}, w_{2} = \frac{- 4 - 52}{2 \cdot 4}

w = \frac{- 4 + 52}{2 \cdot 4} : 6

w = \frac{- 4 - 52}{2 \cdot 4} : - 7

The solutions to the quadratic equation are:

w = 6, w = - 7

(x^{2} - 2 x + 2) = 0

3 x^{2} - 10 x - 5 = 0

x^{2} + 7^{2} = 1 0^{2}

x, x^{2} + 20 x + y^{2} - 20 y + 200 = 0

x, y = 2 - x^{2}

What are the solutions to the equation 4w^2+4w+1=169 ?
The solutions to the equation 4w^2+4w+1=169 are w=6,w=-7
Find the zeros of 4w^2+4w+1=169
The zeros of 4w^2+4w+1=169 are w=6,w=-7