What are the solutions to the equation 2v^2-4v-64=-2 ?

The solutions to the equation 2v^2-4v-64=-2 are v=1+4sqrt(2),v=1-4sqrt(2)

Find the zeros of 2v^2-4v-64=-2

The zeros of 2v^2-4v-64=-2 are v=1+4sqrt(2),v=1-4sqrt(2)

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2 v^{2} - 4 v - 64 = - 2

v = 1 + 42, v = 1 - 42

Decimal Notation

v = 6.65685 \dots, v = - 4.65685 \dots

Solution steps

2 v^{2} - 4 v - 64 = - 2

Move

2

to the left side

2 v^{2} - 4 v - 62 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 ( - 62 )}{2 \cdot 2}

(- 4)^{2} - 4 \cdot 2 (- 62) = 162

v_{1, 2} = \frac{- ( - 4 ) \pm 16 2}{2 \cdot 2}

Separate the solutions

v_{1} = \frac{- ( - 4 ) + 16 2}{2 \cdot 2}, v_{2} = \frac{- ( - 4 ) - 16 2}{2 \cdot 2}

v = \frac{- ( - 4 ) + 16 2}{2 \cdot 2} : 1 + 42

v = \frac{- ( - 4 ) - 16 2}{2 \cdot 2} : 1 - 42

The solutions to the quadratic equation are:

v = 1 + 42, v = 1 - 42

s^{2} + 6 s + 2 = 0

(3 x - 2) (x + 1) = 0

s^{2} + 6 s + 3 = 0

(x + 7)^{2} + (x - 7)^{2} = 130

16 d^{2} = 25

What are the solutions to the equation 2v^2-4v-64=-2 ?
The solutions to the equation 2v^2-4v-64=-2 are v=1+4sqrt(2),v=1-4sqrt(2)
Find the zeros of 2v^2-4v-64=-2
The zeros of 2v^2-4v-64=-2 are v=1+4sqrt(2),v=1-4sqrt(2)