What are the solutions to the equation b^2-3b-28=0 ?

The solutions to the equation b^2-3b-28=0 are b=7,b=-4

Find the zeros of b^2-3b-28=0

The zeros of b^2-3b-28=0 are b=7,b=-4

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b^{2} - 3 b - 28 = 0

b = 7, b = - 4

Solution steps

b^{2} - 3 b - 28 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 28 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 28) = 11

b_{1, 2} = \frac{- ( - 3 ) \pm 11}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- ( - 3 ) + 11}{2 \cdot 1}, b_{2} = \frac{- ( - 3 ) - 11}{2 \cdot 1}

b = \frac{- ( - 3 ) + 11}{2 \cdot 1} : 7

b = \frac{- ( - 3 ) - 11}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

b = 7, b = - 4

28 = x^{2}

m = n - \frac{a x ^{2}}{2}

x^{2} - 12 x - 62 = 0

y + 3 y^{2} = 24

5 v^{2} + 15 v + 8 = 0

What are the solutions to the equation b^2-3b-28=0 ?
The solutions to the equation b^2-3b-28=0 are b=7,b=-4
Find the zeros of b^2-3b-28=0
The zeros of b^2-3b-28=0 are b=7,b=-4