What are the solutions to the equation 4w^2+12w+9=121 ?

The solutions to the equation 4w^2+12w+9=121 are w=4,w=-7

Find the zeros of 4w^2+12w+9=121

The zeros of 4w^2+12w+9=121 are w=4,w=-7

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4 w^{2} + 12 w + 9 = 121

w = 4, w = - 7

Solution steps

4 w^{2} + 12 w + 9 = 121

Move

121

to the left side

4 w^{2} + 12 w - 112 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 4 ( - 112 )}{2 \cdot 4}

1 2^{2} - 4 \cdot 4 (- 112) = 44

w_{1, 2} = \frac{- 12 \pm 44}{2 \cdot 4}

Separate the solutions

w_{1} = \frac{- 12 + 44}{2 \cdot 4}, w_{2} = \frac{- 12 - 44}{2 \cdot 4}

w = \frac{- 12 + 44}{2 \cdot 4} : 4

w = \frac{- 12 - 44}{2 \cdot 4} : - 7

The solutions to the quadratic equation are:

w = 4, w = - 7

3 x + 1 = 5 x - 9

6 x^{2} - 27 x - 15

\frac{120 a b ^{2}}{140 ab}

\frac{16.1 8 ^{2}}{2 \cdot 9.8}

∣ b ∣ = 4

What are the solutions to the equation 4w^2+12w+9=121 ?
The solutions to the equation 4w^2+12w+9=121 are w=4,w=-7
Find the zeros of 4w^2+12w+9=121
The zeros of 4w^2+12w+9=121 are w=4,w=-7