What are the solutions to the equation 4w^2+12w+9=169 ?

The solutions to the equation 4w^2+12w+9=169 are w=5,w=-8

Find the zeros of 4w^2+12w+9=169

The zeros of 4w^2+12w+9=169 are w=5,w=-8

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4 w^{2} + 12 w + 9 = 169

w = 5, w = - 8

Solution steps

4 w^{2} + 12 w + 9 = 169

Move

169

to the left side

4 w^{2} + 12 w - 160 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 4 ( - 160 )}{2 \cdot 4}

1 2^{2} - 4 \cdot 4 (- 160) = 52

w_{1, 2} = \frac{- 12 \pm 52}{2 \cdot 4}

Separate the solutions

w_{1} = \frac{- 12 + 52}{2 \cdot 4}, w_{2} = \frac{- 12 - 52}{2 \cdot 4}

w = \frac{- 12 + 52}{2 \cdot 4} : 5

w = \frac{- 12 - 52}{2 \cdot 4} : - 8

The solutions to the quadratic equation are:

w = 5, w = - 8

5^{2} \cdot 5^{4}

lo g_{3} (\frac{1}{2})

- \frac{m}{8} + 7 \geq 5

- 3 (x + 4) - 2 x = 2 (x + 1)

\frac{( i + 1 )}{( i - 1 )}

What are the solutions to the equation 4w^2+12w+9=169 ?
The solutions to the equation 4w^2+12w+9=169 are w=5,w=-8
Find the zeros of 4w^2+12w+9=169
The zeros of 4w^2+12w+9=169 are w=5,w=-8