What are the solutions to the equation 2b^2+3b=1 ?

The solutions to the equation 2b^2+3b=1 are b=(-3+sqrt(17))/4 ,b=(-3-sqrt(17))/4

Find the zeros of 2b^2+3b=1

The zeros of 2b^2+3b=1 are b=(-3+sqrt(17))/4 ,b=(-3-sqrt(17))/4

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2 b^{2} + 3 b = 1

b = \frac{- 3 + 17}{4}, b = \frac{- 3 - 17}{4}

Decimal Notation

b = 0.28077 \dots, b = - 1.78077 \dots

Solution steps

2 b^{2} + 3 b = 1

Move

1

to the left side

2 b^{2} + 3 b - 1 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

3^{2} - 4 \cdot 2 (- 1) = 17

b_{1, 2} = \frac{- 3 \pm 17}{2 \cdot 2}

Separate the solutions

b_{1} = \frac{- 3 + 17}{2 \cdot 2}, b_{2} = \frac{- 3 - 17}{2 \cdot 2}

b = \frac{- 3 + 17}{2 \cdot 2} : \frac{- 3 + 17}{4}

b = \frac{- 3 - 17}{2 \cdot 2} : \frac{- 3 - 17}{4}

The solutions to the quadratic equation are:

b = \frac{- 3 + 17}{4}, b = \frac{- 3 - 17}{4}

4 m = 25

3 x - 17 = 4

14 x^{2} y - 21 x y + 35 x y^{2}

\frac{2}{3} + \frac{1}{1}

What are the solutions to the equation 2b^2+3b=1 ?
The solutions to the equation 2b^2+3b=1 are b=(-3+sqrt(17))/4 ,b=(-3-sqrt(17))/4
Find the zeros of 2b^2+3b=1
The zeros of 2b^2+3b=1 are b=(-3+sqrt(17))/4 ,b=(-3-sqrt(17))/4