What are the solutions to the equation 4b^2-19b+20=0 ?

The solutions to the equation 4b^2-19b+20=0 are b=(19+sqrt(41))/8 ,b=(19-sqrt(41))/8

Find the zeros of 4b^2-19b+20=0

The zeros of 4b^2-19b+20=0 are b=(19+sqrt(41))/8 ,b=(19-sqrt(41))/8

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4 b^{2} - 19 b + 20 = 0

b = \frac{19 + 41}{8}, b = \frac{19 - 41}{8}

Decimal Notation

b = 3.17539 \dots, b = 1.57460 \dots

Solution steps

4 b^{2} - 19 b + 20 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- ( - 19 ) \pm ( - 19 ) ^{2} - 4 \cdot 4 \cdot 20}{2 \cdot 4}

(- 19)^{2} - 4 \cdot 4 \cdot 20 = 41

b_{1, 2} = \frac{- ( - 19 ) \pm 41}{2 \cdot 4}

Separate the solutions

b_{1} = \frac{- ( - 19 ) + 41}{2 \cdot 4}, b_{2} = \frac{- ( - 19 ) - 41}{2 \cdot 4}

b = \frac{- ( - 19 ) + 41}{2 \cdot 4} : \frac{19 + 41}{8}

b = \frac{- ( - 19 ) - 41}{2 \cdot 4} : \frac{19 - 41}{8}

The solutions to the quadratic equation are:

b = \frac{19 + 41}{8}, b = \frac{19 - 41}{8}

6 x (x - 1) = 4 x + 5

(\frac{1}{8} - x)^{2} + 42 = (x + \frac{1}{8})^{2} - \frac{x}{2} + x (x - 1)

2 q^{2} - 3 q - 15 = - 4 q

11 d^{2} - 12 d + 4 = 6 d^{2}

- x^{2} = 8 x - 2

What are the solutions to the equation 4b^2-19b+20=0 ?
The solutions to the equation 4b^2-19b+20=0 are b=(19+sqrt(41))/8 ,b=(19-sqrt(41))/8
Find the zeros of 4b^2-19b+20=0
The zeros of 4b^2-19b+20=0 are b=(19+sqrt(41))/8 ,b=(19-sqrt(41))/8