What are the solutions to the equation 2b^2+12b+15=0 ?

The solutions to the equation 2b^2+12b+15=0 are b=(-6+sqrt(6))/2 ,b=-(3sqrt(2)+sqrt(3))/(sqrt(2))

Find the zeros of 2b^2+12b+15=0

The zeros of 2b^2+12b+15=0 are b=(-6+sqrt(6))/2 ,b=-(3sqrt(2)+sqrt(3))/(sqrt(2))

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2 b^{2} + 12 b + 15 = 0

b = \frac{- 6 + 6}{2}, b = - \frac{3 2 + 3}{2}

Decimal Notation

b = - 1.77525 \dots, b = - 4.22474 \dots

Solution steps

2 b^{2} + 12 b + 15 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 2 \cdot 15}{2 \cdot 2}

1 2^{2} - 4 \cdot 2 \cdot 15 = 26

b_{1, 2} = \frac{- 12 \pm 2 6}{2 \cdot 2}

Separate the solutions

b_{1} = \frac{- 12 + 2 6}{2 \cdot 2}, b_{2} = \frac{- 12 - 2 6}{2 \cdot 2}

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

b = \frac{- 12 - 2 6}{2 \cdot 2} : - \frac{3 2 + 3}{2}

The solutions to the quadratic equation are:

b = \frac{- 6 + 6}{2}, b = - \frac{3 2 + 3}{2}

(x + 3) (x - 4) < - 12

x^{3} - 4 x^{2} + 3 x - 12

24 3^{\frac{6}{5}}

7^{\frac{1}{3}} \cdot 7^{\frac{4}{3}}

∣ 3 x - 4 ∣ < 1

What are the solutions to the equation 2b^2+12b+15=0 ?
The solutions to the equation 2b^2+12b+15=0 are b=(-6+sqrt(6))/2 ,b=-(3sqrt(2)+sqrt(3))/(sqrt(2))
Find the zeros of 2b^2+12b+15=0
The zeros of 2b^2+12b+15=0 are b=(-6+sqrt(6))/2 ,b=-(3sqrt(2)+sqrt(3))/(sqrt(2))