What are the solutions to the equation v^2-17v+16=-2v^2 ?

The solutions to the equation v^2-17v+16=-2v^2 are v=(17+sqrt(97))/6 ,v=(17-sqrt(97))/6

Find the zeros of v^2-17v+16=-2v^2

The zeros of v^2-17v+16=-2v^2 are v=(17+sqrt(97))/6 ,v=(17-sqrt(97))/6

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v^{2} - 17 v + 16 = - 2 v^{2}

v = \frac{17 + 97}{6}, v = \frac{17 - 97}{6}

Decimal Notation

v = 4.47480 \dots, v = 1.19185 \dots

Solution steps

v^{2} - 17 v + 16 = - 2 v^{2}

Move

2 v^{2}

to the left side

3 v^{2} - 17 v + 16 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 \cdot 3 \cdot 16}{2 \cdot 3}

(- 17)^{2} - 4 \cdot 3 \cdot 16 = 97

v_{1, 2} = \frac{- ( - 17 ) \pm 97}{2 \cdot 3}

Separate the solutions

v_{1} = \frac{- ( - 17 ) + 97}{2 \cdot 3}, v_{2} = \frac{- ( - 17 ) - 97}{2 \cdot 3}

v = \frac{- ( - 17 ) + 97}{2 \cdot 3} : \frac{17 + 97}{6}

v = \frac{- ( - 17 ) - 97}{2 \cdot 3} : \frac{17 - 97}{6}

The solutions to the quadratic equation are:

v = \frac{17 + 97}{6}, v = \frac{17 - 97}{6}

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What are the solutions to the equation v^2-17v+16=-2v^2 ?
The solutions to the equation v^2-17v+16=-2v^2 are v=(17+sqrt(97))/6 ,v=(17-sqrt(97))/6
Find the zeros of v^2-17v+16=-2v^2
The zeros of v^2-17v+16=-2v^2 are v=(17+sqrt(97))/6 ,v=(17-sqrt(97))/6