What are the solutions to the equation 4b^2+11b=5 ?

The solutions to the equation 4b^2+11b=5 are b=(-11+sqrt(201))/8 ,b=(-11-sqrt(201))/8

Find the zeros of 4b^2+11b=5

The zeros of 4b^2+11b=5 are b=(-11+sqrt(201))/8 ,b=(-11-sqrt(201))/8

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4 b^{2} + 11 b = 5

b = \frac{- 11 + 201}{8}, b = \frac{- 11 - 201}{8}

Decimal Notation

b = 0.39718 \dots, b = - 3.14718 \dots

Solution steps

4 b^{2} + 11 b = 5

Move

5

to the left side

4 b^{2} + 11 b - 5 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 4 ( - 5 )}{2 \cdot 4}

1 1^{2} - 4 \cdot 4 (- 5) = 201

b_{1, 2} = \frac{- 11 \pm 201}{2 \cdot 4}

Separate the solutions

b_{1} = \frac{- 11 + 201}{2 \cdot 4}, b_{2} = \frac{- 11 - 201}{2 \cdot 4}

b = \frac{- 11 + 201}{2 \cdot 4} : \frac{- 11 + 201}{8}

b = \frac{- 11 - 201}{2 \cdot 4} : \frac{- 11 - 201}{8}

The solutions to the quadratic equation are:

b = \frac{- 11 + 201}{8}, b = \frac{- 11 - 201}{8}

6 x > - 12

14 x^{2} + 25 x + 6

4 x^{2} + 4 x - 8

3 x + 18 + x y + 6 y

\frac{- ( \frac{4}{3} ) ^{2} + 2 ( \frac{8}{3} )}{( \frac{8}{3} ) ^{2} - 2 ( \frac{4}{3} )}

What are the solutions to the equation 4b^2+11b=5 ?
The solutions to the equation 4b^2+11b=5 are b=(-11+sqrt(201))/8 ,b=(-11-sqrt(201))/8
Find the zeros of 4b^2+11b=5
The zeros of 4b^2+11b=5 are b=(-11+sqrt(201))/8 ,b=(-11-sqrt(201))/8