What are the solutions to the equation 20=b^2-b ?

The solutions to the equation 20=b^2-b are b=5,b=-4

Find the zeros of 20=b^2-b

The zeros of 20=b^2-b are b=5,b=-4

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20 = b^{2} - b

b = 5, b = - 4

Solution steps

20 = b^{2} - b

Switch sides

b^{2} - b = 20

Move

20

to the left side

b^{2} - b - 20 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 20 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 20) = 9

b_{1, 2} = \frac{- ( - 1 ) \pm 9}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- ( - 1 ) + 9}{2 \cdot 1}, b_{2} = \frac{- ( - 1 ) - 9}{2 \cdot 1}

b = \frac{- ( - 1 ) + 9}{2 \cdot 1} : 5

b = \frac{- ( - 1 ) - 9}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

b = 5, b = - 4

8 x^{5} - 14 x^{4} - 22 x^{3} + 57 x^{2} - 35 x + 6

(- 1)^{2}

\frac{7}{12} + \frac{1}{3}

(- 216)^{\frac{2}{3}}

x^{2} + 8 x = 4

What are the solutions to the equation 20=b^2-b ?
The solutions to the equation 20=b^2-b are b=5,b=-4
Find the zeros of 20=b^2-b
The zeros of 20=b^2-b are b=5,b=-4