What are the solutions to the equation 2b^2-5b+7=0 ?

The solutions to the equation 2b^2-5b+7=0 are b= 5/4+i(sqrt(31))/4 ,b= 5/4-i(sqrt(31))/4

Find the zeros of 2b^2-5b+7=0

The zeros of 2b^2-5b+7=0 are b= 5/4+i(sqrt(31))/4 ,b= 5/4-i(sqrt(31))/4

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2 b^{2} - 5 b + 7 = 0

b = \frac{5}{4} + i \frac{31}{4}, b = \frac{5}{4} - i \frac{31}{4}

Solution steps

2 b^{2} - 5 b + 7 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 \cdot 7}{2 \cdot 2}

Simplify

(- 5)^{2} - 4 \cdot 2 \cdot 7 : 31 i

b_{1, 2} = \frac{- ( - 5 ) \pm 31 i}{2 \cdot 2}

Separate the solutions

b_{1} = \frac{- ( - 5 ) + 31 i}{2 \cdot 2}, b_{2} = \frac{- ( - 5 ) - 31 i}{2 \cdot 2}

b = \frac{- ( - 5 ) + 31 i}{2 \cdot 2} : \frac{5}{4} + i \frac{31}{4}

b = \frac{- ( - 5 ) - 31 i}{2 \cdot 2} : \frac{5}{4} - i \frac{31}{4}

The solutions to the quadratic equation are:

b = \frac{5}{4} + i \frac{31}{4}, b = \frac{5}{4} - i \frac{31}{4}

- 3 v - 14.47 > - 5.47

50 v^{2} + 40 v d + 8 d^{2}

4 x^{2} - x

lo g_{10} (2)

- 3 (0) + y = - 4

What are the solutions to the equation 2b^2-5b+7=0 ?
The solutions to the equation 2b^2-5b+7=0 are b= 5/4+i(sqrt(31))/4 ,b= 5/4-i(sqrt(31))/4
Find the zeros of 2b^2-5b+7=0
The zeros of 2b^2-5b+7=0 are b= 5/4+i(sqrt(31))/4 ,b= 5/4-i(sqrt(31))/4