What are the solutions to the equation 3b^2-b+5=6 ?

The solutions to the equation 3b^2-b+5=6 are b=(1+sqrt(13))/6 ,b=(1-sqrt(13))/6

Find the zeros of 3b^2-b+5=6

The zeros of 3b^2-b+5=6 are b=(1+sqrt(13))/6 ,b=(1-sqrt(13))/6

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3 b^{2} - b + 5 = 6

b = \frac{1 + 13}{6}, b = \frac{1 - 13}{6}

Solution steps

3 b^{2} - b + 5 = 6

Move

6

to the left side

3 b^{2} - b - 1 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 3 (- 1) = 13

b_{1, 2} = \frac{- ( - 1 ) \pm 13}{2 \cdot 3}

Separate the solutions

b_{1} = \frac{- ( - 1 ) + 13}{2 \cdot 3}, b_{2} = \frac{- ( - 1 ) - 13}{2 \cdot 3}

b = \frac{- ( - 1 ) + 13}{2 \cdot 3} : \frac{1 + 13}{6}

b = \frac{- ( - 1 ) - 13}{2 \cdot 3} : \frac{1 - 13}{6}

The solutions to the quadratic equation are:

b = \frac{1 + 13}{6}, b = \frac{1 - 13}{6}

x, (2 x - 9 y)^{2} + 7 y = 27 x

3 y^{2} + 2 = 5 y

\frac{( 2 x + 3 ) ^{2}}{2} = - 1

9 x^{2} - 8 = 0

x, x^{2} y^{2} + x y = 1

What are the solutions to the equation 3b^2-b+5=6 ?
The solutions to the equation 3b^2-b+5=6 are b=(1+sqrt(13))/6 ,b=(1-sqrt(13))/6
Find the zeros of 3b^2-b+5=6
The zeros of 3b^2-b+5=6 are b=(1+sqrt(13))/6 ,b=(1-sqrt(13))/6