What are the solutions to the equation v^2-16v-32=0 ?

The solutions to the equation v^2-16v-32=0 are v=4(2+sqrt(6)),v=4(2-sqrt(6))

Find the zeros of v^2-16v-32=0

The zeros of v^2-16v-32=0 are v=4(2+sqrt(6)),v=4(2-sqrt(6))

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v^{2} - 16 v - 32 = 0

v = 4 (2 + 6), v = 4 (2 - 6)

Decimal Notation

v = 17.79795 \dots, v = - 1.79795 \dots

Solution steps

v^{2} - 16 v - 32 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 1 \cdot ( - 32 )}{2 \cdot 1}

(- 16)^{2} - 4 \cdot 1 \cdot (- 32) = 86

v_{1, 2} = \frac{- ( - 16 ) \pm 8 6}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( - 16 ) + 8 6}{2 \cdot 1}, v_{2} = \frac{- ( - 16 ) - 8 6}{2 \cdot 1}

v = \frac{- ( - 16 ) + 8 6}{2 \cdot 1} : 4 (2 + 6)

v = \frac{- ( - 16 ) - 8 6}{2 \cdot 1} : 4 (2 - 6)

The solutions to the quadratic equation are:

v = 4 (2 + 6), v = 4 (2 - 6)

(x + 4)^{2} + (x - 7)^{2} = 25

- 7 x^{2} - 4 = 325

x (x + 3) - 10 = 0

(- 4 - m)^{2} - 4 = 0

6 x^{2} + 3 x - 14 = 0

What are the solutions to the equation v^2-16v-32=0 ?
The solutions to the equation v^2-16v-32=0 are v=4(2+sqrt(6)),v=4(2-sqrt(6))
Find the zeros of v^2-16v-32=0
The zeros of v^2-16v-32=0 are v=4(2+sqrt(6)),v=4(2-sqrt(6))