What are the solutions to the equation b^2-12b-56=0 ?

The solutions to the equation b^2-12b-56=0 are b=2(3+sqrt(23)),b=2(3-sqrt(23))

Find the zeros of b^2-12b-56=0

The zeros of b^2-12b-56=0 are b=2(3+sqrt(23)),b=2(3-sqrt(23))

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b^{2} - 12 b - 56 = 0

b = 2 (3 + 23), b = 2 (3 - 23)

Decimal Notation

b = 15.59166 \dots, b = - 3.59166 \dots

Solution steps

b^{2} - 12 b - 56 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot ( - 56 )}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot (- 56) = 423

b_{1, 2} = \frac{- ( - 12 ) \pm 4 23}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- ( - 12 ) + 4 23}{2 \cdot 1}, b_{2} = \frac{- ( - 12 ) - 4 23}{2 \cdot 1}

b = \frac{- ( - 12 ) + 4 23}{2 \cdot 1} : 2 (3 + 23)

b = \frac{- ( - 12 ) - 4 23}{2 \cdot 1} : 2 (3 - 23)

The solutions to the quadratic equation are:

b = 2 (3 + 23), b = 2 (3 - 23)

6 (\frac{3}{2})

4^{2} + 5^{2} = 6^{2}

6 x - 15 > - 45

10 x^{3} - 20 x^{2} - 2 x + 4

x^{3} - 7 x^{2} + 24 x - 18

What are the solutions to the equation b^2-12b-56=0 ?
The solutions to the equation b^2-12b-56=0 are b=2(3+sqrt(23)),b=2(3-sqrt(23))
Find the zeros of b^2-12b-56=0
The zeros of b^2-12b-56=0 are b=2(3+sqrt(23)),b=2(3-sqrt(23))