What are the solutions to the equation w^2+12w=71 ?

The solutions to the equation w^2+12w=71 are w=-6+sqrt(107),w=-6-sqrt(107)

Find the zeros of w^2+12w=71

The zeros of w^2+12w=71 are w=-6+sqrt(107),w=-6-sqrt(107)

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w^{2} + 12 w = 71

w = - 6 + 107, w = - 6 - 107

Decimal Notation

w = 4.34408 \dots, w = - 16.34408 \dots

Solution steps

w^{2} + 12 w = 71

Move

71

to the left side

w^{2} + 12 w - 71 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot ( - 71 )}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot (- 71) = 2107

w_{1, 2} = \frac{- 12 \pm 2 107}{2 \cdot 1}

Separate the solutions

w_{1} = \frac{- 12 + 2 107}{2 \cdot 1}, w_{2} = \frac{- 12 - 2 107}{2 \cdot 1}

w = \frac{- 12 + 2 107}{2 \cdot 1} : - 6 + 107

w = \frac{- 12 - 2 107}{2 \cdot 1} : - 6 - 107

The solutions to the quadratic equation are:

w = - 6 + 107, w = - 6 - 107

\frac{1 0 ^{5}}{1 0 ^{6}}

x^{2} - 4 x + 1 = 6

5 x - 3 \leq 7 - 3 x

x^{3} y^{3} (3 y^{3})

2 < 2 x + 2 \leq 8

What are the solutions to the equation w^2+12w=71 ?
The solutions to the equation w^2+12w=71 are w=-6+sqrt(107),w=-6-sqrt(107)
Find the zeros of w^2+12w=71
The zeros of w^2+12w=71 are w=-6+sqrt(107),w=-6-sqrt(107)