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

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

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

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

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

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

Decimal Notation

c = 4.47480 \dots, c = 1.19185 \dots

Solution steps

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

Move

2 c^{2}

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

2 v^{2} + 3 v - 5 = - 1

2 w (w - 1) = 5

7 y^{2} - y - 4 = 6 y^{2}

0 = x^{2} - 0.3 x - 0.1

a, a^{2} + b^{2} + c^{2} = 2015

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