What are the solutions to the equation b^2-9b+20=0 ?

The solutions to the equation b^2-9b+20=0 are b=5,b=4

Find the zeros of b^2-9b+20=0

The zeros of b^2-9b+20=0 are b=5,b=4

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b^{2} - 9 b + 20 = 0

b = 5, b = 4

Solution steps

b^{2} - 9 b + 20 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot 20}{2 \cdot 1}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

b = 5, b = 4

49 x^{2} = 121

x^{2} - 5 x - 3 = 0

x^{2} - 9 x + 20 = 0

s^{2} = - 4

3 x^{2} - 3 x - 20 = 2 x + 8

What are the solutions to the equation b^2-9b+20=0 ?
The solutions to the equation b^2-9b+20=0 are b=5,b=4
Find the zeros of b^2-9b+20=0
The zeros of b^2-9b+20=0 are b=5,b=4