What are the solutions to the equation w^2-12w-12=0 ?

The solutions to the equation w^2-12w-12=0 are w=2(3+2sqrt(3)),w=2(3-2sqrt(3))

Find the zeros of w^2-12w-12=0

The zeros of w^2-12w-12=0 are w=2(3+2sqrt(3)),w=2(3-2sqrt(3))

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w^{2} - 12 w - 12 = 0

w = 2 (3 + 23), w = 2 (3 - 23)

Decimal Notation

w = 12.92820 \dots, w = - 0.92820 \dots

Solution steps

w^{2} - 12 w - 12 = 0

Solve with the quadratic formula

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

(- 12)^{2} - 4 \cdot 1 \cdot (- 12) = 83

w_{1, 2} = \frac{- ( - 12 ) \pm 8 3}{2 \cdot 1}

Separate the solutions

w_{1} = \frac{- ( - 12 ) + 8 3}{2 \cdot 1}, w_{2} = \frac{- ( - 12 ) - 8 3}{2 \cdot 1}

w = \frac{- ( - 12 ) + 8 3}{2 \cdot 1} : 2 (3 + 23)

w = \frac{- ( - 12 ) - 8 3}{2 \cdot 1} : 2 (3 - 23)

The solutions to the quadratic equation are:

w = 2 (3 + 23), w = 2 (3 - 23)

t = 3 t^{2} + 2 t, \int_{0}^{2} t d t

5 x^{2} + 4 x - 10 = 0

x^{2} - 16 x + 35 = 0

x^{2} + (x + 2)^{2} = 10

\frac{5}{6} x - \frac{2}{9} x^{2} = 0

What are the solutions to the equation w^2-12w-12=0 ?
The solutions to the equation w^2-12w-12=0 are w=2(3+2sqrt(3)),w=2(3-2sqrt(3))
Find the zeros of w^2-12w-12=0
The zeros of w^2-12w-12=0 are w=2(3+2sqrt(3)),w=2(3-2sqrt(3))