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

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

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

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

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

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

Decimal Notation

b = 4.47480 \dots, b = 1.19185 \dots

Solution steps

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

Move

2 b^{2}

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

5 x^{2} + 13 x + 9 = 3

4 x^{2} + 23 = 19 x

60 x^{2} - 14 = 0

2 q^{2} - 44 q = - 14

4 - 36 x^{2} = 0

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