What are the solutions to the equation 2b^2+4b-126=0 ?

The solutions to the equation 2b^2+4b-126=0 are b=7,b=-9

Find the zeros of 2b^2+4b-126=0

The zeros of 2b^2+4b-126=0 are b=7,b=-9

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2 b^{2} + 4 b - 126 = 0

b = 7, b = - 9

Solution steps

2 b^{2} + 4 b - 126 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 ( - 126 )}{2 \cdot 2}

4^{2} - 4 \cdot 2 (- 126) = 32

b_{1, 2} = \frac{- 4 \pm 32}{2 \cdot 2}

Separate the solutions

b_{1} = \frac{- 4 + 32}{2 \cdot 2}, b_{2} = \frac{- 4 - 32}{2 \cdot 2}

b = \frac{- 4 + 32}{2 \cdot 2} : 7

b = \frac{- 4 - 32}{2 \cdot 2} : - 9

The solutions to the quadratic equation are:

b = 7, b = - 9

z^{2} + 0.2 z - 0.08 = 0

x^{2} = x + 30

- 3 x^{2} + 15 = 0

15 q^{2} + 31 q + 15 = 5

(2 x + 1) (x + 1) + (2 x - 1) x = 7

What are the solutions to the equation 2b^2+4b-126=0 ?
The solutions to the equation 2b^2+4b-126=0 are b=7,b=-9
Find the zeros of 2b^2+4b-126=0
The zeros of 2b^2+4b-126=0 are b=7,b=-9