What are the solutions to the equation (2b^2+8b)-19=5 ?

The solutions to the equation (2b^2+8b)-19=5 are b=2,b=-6

Find the zeros of (2b^2+8b)-19=5

The zeros of (2b^2+8b)-19=5 are b=2,b=-6

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(2 b^{2} + 8 b) - 19 = 5

b = 2, b = - 6

Solution steps

(2 b^{2} + 8 b) - 19 = 5

Refine

2 b^{2} + 8 b - 19 = 5

Move

5

to the left side

2 b^{2} + 8 b - 24 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 2 ( - 24 )}{2 \cdot 2}

8^{2} - 4 \cdot 2 (- 24) = 16

b_{1, 2} = \frac{- 8 \pm 16}{2 \cdot 2}

Separate the solutions

b_{1} = \frac{- 8 + 16}{2 \cdot 2}, b_{2} = \frac{- 8 - 16}{2 \cdot 2}

b = \frac{- 8 + 16}{2 \cdot 2} : 2

b = \frac{- 8 - 16}{2 \cdot 2} : - 6

The solutions to the quadratic equation are:

b = 2, b = - 6

x^{2} = 4 b

- 2 x^{2} - 4 x + 0

3 x^{2} - 32 = 0

- 53 = 1 - 6 n^{2}

x^{2} - 13 x = 90

What are the solutions to the equation (2b^2+8b)-19=5 ?
The solutions to the equation (2b^2+8b)-19=5 are b=2,b=-6
Find the zeros of (2b^2+8b)-19=5
The zeros of (2b^2+8b)-19=5 are b=2,b=-6