What are the solutions to the equation w^2+5w-86=0 ?

The solutions to the equation w^2+5w-86=0 are w=(-5+3sqrt(41))/2 ,w=(-5-3sqrt(41))/2

Find the zeros of w^2+5w-86=0

The zeros of w^2+5w-86=0 are w=(-5+3sqrt(41))/2 ,w=(-5-3sqrt(41))/2

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w^{2} + 5 w - 86 = 0

w = \frac{- 5 + 3 41}{2}, w = \frac{- 5 - 3 41}{2}

Decimal Notation

w = 7.10468 \dots, w = - 12.10468 \dots

Solution steps

w^{2} + 5 w - 86 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot ( - 86 )}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot (- 86) = 341

w_{1, 2} = \frac{- 5 \pm 3 41}{2 \cdot 1}

Separate the solutions

w_{1} = \frac{- 5 + 3 41}{2 \cdot 1}, w_{2} = \frac{- 5 - 3 41}{2 \cdot 1}

w = \frac{- 5 + 3 41}{2 \cdot 1} : \frac{- 5 + 3 41}{2}

w = \frac{- 5 - 3 41}{2 \cdot 1} : \frac{- 5 - 3 41}{2}

The solutions to the quadratic equation are:

w = \frac{- 5 + 3 41}{2}, w = \frac{- 5 - 3 41}{2}

∣ 2 x - 6 ∣ \leq 12

5 x^{2} + 57 x - 36 = 0

∣ 9 x + 2 ∣ + 4 < 6

\frac{( - 4 ) ^{8} \cdot ( - 4 ) ^{3}}{( - 4 ) ^{4} \cdot ( - 4 ) ^{2}}

\frac{- 25}{- 5}

What are the solutions to the equation w^2+5w-86=0 ?
The solutions to the equation w^2+5w-86=0 are w=(-5+3sqrt(41))/2 ,w=(-5-3sqrt(41))/2
Find the zeros of w^2+5w-86=0
The zeros of w^2+5w-86=0 are w=(-5+3sqrt(41))/2 ,w=(-5-3sqrt(41))/2