What are the solutions to the equation -2b^2-4b-20=-3b^2 ?

The solutions to the equation -2b^2-4b-20=-3b^2 are b=2(1+sqrt(6)),b=2(1-sqrt(6))

Find the zeros of -2b^2-4b-20=-3b^2

The zeros of -2b^2-4b-20=-3b^2 are b=2(1+sqrt(6)),b=2(1-sqrt(6))

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- 2 b^{2} - 4 b - 20 = - 3 b^{2}

b = 2 (1 + 6), b = 2 (1 - 6)

Decimal Notation

b = 6.89897 \dots, b = - 2.89897 \dots

Solution steps

- 2 b^{2} - 4 b - 20 = - 3 b^{2}

Move

3 b^{2}

to the left side

b^{2} - 4 b - 20 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 20 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 20) = 46

b_{1, 2} = \frac{- ( - 4 ) \pm 4 6}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- ( - 4 ) + 4 6}{2 \cdot 1}, b_{2} = \frac{- ( - 4 ) - 4 6}{2 \cdot 1}

b = \frac{- ( - 4 ) + 4 6}{2 \cdot 1} : 2 (1 + 6)

b = \frac{- ( - 4 ) - 4 6}{2 \cdot 1} : 2 (1 - 6)

The solutions to the quadratic equation are:

b = 2 (1 + 6), b = 2 (1 - 6)

x^{2} - 6 x = 0

x^{2} + 6 x + 9 = 49

3 p^{2} + 12 p = 6

x^{2} - 20 x + 28 = 0

5^{2} = 3^{2} + x^{2}

What are the solutions to the equation -2b^2-4b-20=-3b^2 ?
The solutions to the equation -2b^2-4b-20=-3b^2 are b=2(1+sqrt(6)),b=2(1-sqrt(6))
Find the zeros of -2b^2-4b-20=-3b^2
The zeros of -2b^2-4b-20=-3b^2 are b=2(1+sqrt(6)),b=2(1-sqrt(6))