What are the solutions to the equation v^2=54-12v ?

The solutions to the equation v^2=54-12v are v=3(sqrt(10)-2),v=-3(2+sqrt(10))

Find the zeros of v^2=54-12v

The zeros of v^2=54-12v are v=3(sqrt(10)-2),v=-3(2+sqrt(10))

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v^{2} = 54 - 12 v

v = 3 (10 - 2), v = - 3 (2 + 10)

Solution steps

v^{2} = 54 - 12 v

Move

12 v

to the left side

v^{2} + 12 v = 54

Move

54

to the left side

v^{2} + 12 v - 54 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot ( - 54 )}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot (- 54) = 610

v_{1, 2} = \frac{- 12 \pm 6 10}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- 12 + 6 10}{2 \cdot 1}, v_{2} = \frac{- 12 - 6 10}{2 \cdot 1}

v = \frac{- 12 + 6 10}{2 \cdot 1} : 3 (10 - 2)

v = \frac{- 12 - 6 10}{2 \cdot 1} : - 3 (2 + 10)

The solutions to the quadratic equation are:

v = 3 (10 - 2), v = - 3 (2 + 10)

\frac{3}{7} x + 5 = 8

36 g^{2} - 49 h^{2}

x^{2} - x - 20 \leq 0

5 x - 10 = 3 x^{2}

(8 \cdot 320)^{\frac{1}{3}}

What are the solutions to the equation v^2=54-12v ?
The solutions to the equation v^2=54-12v are v=3(sqrt(10)-2),v=-3(2+sqrt(10))
Find the zeros of v^2=54-12v
The zeros of v^2=54-12v are v=3(sqrt(10)-2),v=-3(2+sqrt(10))