What are the solutions to the equation b^2-b-20=0 ?

The solutions to the equation b^2-b-20=0 are b=5,b=-4

Find the zeros of b^2-b-20=0

The zeros of b^2-b-20=0 are b=5,b=-4

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b^{2} - b - 20 = 0

b = 5, b = - 4

Solution steps

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

(z + 8)^{2} = - 15

22 > 4 m - 2

(\frac{2 + i}{3 + 4 i} - \frac{2 i}{3 - 4 i})^{2}

\frac{1}{3} (x + 3) + \frac{1}{2} \geq - \frac{9}{2}

2 (1 - x) \geq 5 x

What are the solutions to the equation b^2-b-20=0 ?
The solutions to the equation b^2-b-20=0 are b=5,b=-4
Find the zeros of b^2-b-20=0
The zeros of b^2-b-20=0 are b=5,b=-4