What are the solutions to the equation b^2+b=42 ?

The solutions to the equation b^2+b=42 are b=6,b=-7

Find the zeros of b^2+b=42

The zeros of b^2+b=42 are b=6,b=-7

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b^{2} + b = 42

b = 6, b = - 7

Solution steps

b^{2} + b = 42

Move

42

to the left side

b^{2} + b - 42 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 42 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 42) = 13

b_{1, 2} = \frac{- 1 \pm 13}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- 1 + 13}{2 \cdot 1}, b_{2} = \frac{- 1 - 13}{2 \cdot 1}

b = \frac{- 1 + 13}{2 \cdot 1} : 6

b = \frac{- 1 - 13}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

b = 6, b = - 7

3 y^{2} + 5 y + 4 = 2

4^{2} = 2 x^{2}

v^{2} = - 16 - 10 v

- (x - 4) (2 x + 3) = 2 (4 x + 5)

32 = 16 + 32 t - 16 t^{2}

What are the solutions to the equation b^2+b=42 ?
The solutions to the equation b^2+b=42 are b=6,b=-7
Find the zeros of b^2+b=42
The zeros of b^2+b=42 are b=6,b=-7