What are the solutions to the equation 2b^2+12b+23=6 ?

The solutions to the equation 2b^2+12b+23=6 are b=(-6+sqrt(2))/2 ,b=-(6+sqrt(2))/2

Find the zeros of 2b^2+12b+23=6

The zeros of 2b^2+12b+23=6 are b=(-6+sqrt(2))/2 ,b=-(6+sqrt(2))/2

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2 b^{2} + 12 b + 23 = 6

b = \frac{- 6 + 2}{2}, b = - \frac{6 + 2}{2}

Decimal Notation

b = - 2.29289 \dots, b = - 3.70710 \dots

Solution steps

2 b^{2} + 12 b + 23 = 6

Move

6

to the left side

2 b^{2} + 12 b + 17 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 2 \cdot 17}{2 \cdot 2}

1 2^{2} - 4 \cdot 2 \cdot 17 = 22

b_{1, 2} = \frac{- 12 \pm 2 2}{2 \cdot 2}

Separate the solutions

b_{1} = \frac{- 12 + 2 2}{2 \cdot 2}, b_{2} = \frac{- 12 - 2 2}{2 \cdot 2}

b = \frac{- 12 + 2 2}{2 \cdot 2} : \frac{- 6 + 2}{2}

b = \frac{- 12 - 2 2}{2 \cdot 2} : - \frac{6 + 2}{2}

The solutions to the quadratic equation are:

b = \frac{- 6 + 2}{2}, b = - \frac{6 + 2}{2}

x = 2 x^{2} - 2 x + 3

25 x^{2} - 225 x = 0

12 y^{2} - 9 y - 3 = 0

3 (2 x - 6)^{2} + 6 = 27

12 a^{2} - 22 a - 5 = - 5 a

What are the solutions to the equation 2b^2+12b+23=6 ?
The solutions to the equation 2b^2+12b+23=6 are b=(-6+sqrt(2))/2 ,b=-(6+sqrt(2))/2
Find the zeros of 2b^2+12b+23=6
The zeros of 2b^2+12b+23=6 are b=(-6+sqrt(2))/2 ,b=-(6+sqrt(2))/2