What are the solutions to the equation v^2-16v-48=-4 ?

The solutions to the equation v^2-16v-48=-4 are v=2(4+3sqrt(3)),v=2(4-3sqrt(3))

Find the zeros of v^2-16v-48=-4

The zeros of v^2-16v-48=-4 are v=2(4+3sqrt(3)),v=2(4-3sqrt(3))

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v^{2} - 16 v - 48 = - 4

v = 2 (4 + 33), v = 2 (4 - 33)

Decimal Notation

v = 18.39230 \dots, v = - 2.39230 \dots

Solution steps

v^{2} - 16 v - 48 = - 4

Move

4

to the left side

v^{2} - 16 v - 44 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 1 \cdot ( - 44 )}{2 \cdot 1}

(- 16)^{2} - 4 \cdot 1 \cdot (- 44) = 123

v_{1, 2} = \frac{- ( - 16 ) \pm 12 3}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( - 16 ) + 12 3}{2 \cdot 1}, v_{2} = \frac{- ( - 16 ) - 12 3}{2 \cdot 1}

v = \frac{- ( - 16 ) + 12 3}{2 \cdot 1} : 2 (4 + 33)

v = \frac{- ( - 16 ) - 12 3}{2 \cdot 1} : 2 (4 - 33)

The solutions to the quadratic equation are:

v = 2 (4 + 33), v = 2 (4 - 33)

- x^{2} + 100 x = 0

2 b^{2} + 5 b = 42

y^{2} + 10 y + 9 = 0

t, d = 16 t^{2}

y, x^{2} + 2 x y + y^{2} = 1

What are the solutions to the equation v^2-16v-48=-4 ?
The solutions to the equation v^2-16v-48=-4 are v=2(4+3sqrt(3)),v=2(4-3sqrt(3))
Find the zeros of v^2-16v-48=-4
The zeros of v^2-16v-48=-4 are v=2(4+3sqrt(3)),v=2(4-3sqrt(3))