What are the solutions to the equation 2v^2+12=14v ?

The solutions to the equation 2v^2+12=14v are v=6,v=1

Find the zeros of 2v^2+12=14v

The zeros of 2v^2+12=14v are v=6,v=1

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2 v^{2} + 12 = 14 v

v = 6, v = 1

Solution steps

2 v^{2} + 12 = 14 v

Move

14 v

to the left side

2 v^{2} + 12 - 14 v = 0

Write in the standard form

a x^{2} + b x + c = 0

2 v^{2} - 14 v + 12 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 2 \cdot 12}{2 \cdot 2}

(- 14)^{2} - 4 \cdot 2 \cdot 12 = 10

v_{1, 2} = \frac{- ( - 14 ) \pm 10}{2 \cdot 2}

Separate the solutions

v_{1} = \frac{- ( - 14 ) + 10}{2 \cdot 2}, v_{2} = \frac{- ( - 14 ) - 10}{2 \cdot 2}

v = \frac{- ( - 14 ) + 10}{2 \cdot 2} : 6

v = \frac{- ( - 14 ) - 10}{2 \cdot 2} : 1

The solutions to the quadratic equation are:

v = 6, v = 1

(x^{2} - 8) = 0

\frac{1}{2} - y^{2} + 3 = 0

a^{2} - 4 a - 5 = 0

9 x^{2} + 5 = 0

(2 x + 10)^{2} = 100 (1 - x)

What are the solutions to the equation 2v^2+12=14v ?
The solutions to the equation 2v^2+12=14v are v=6,v=1
Find the zeros of 2v^2+12=14v
The zeros of 2v^2+12=14v are v=6,v=1