What are the solutions to the equation 2b^2+5b=42 ?

The solutions to the equation 2b^2+5b=42 are b= 7/2 ,b=-6

Find the zeros of 2b^2+5b=42

The zeros of 2b^2+5b=42 are b= 7/2 ,b=-6

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2 b^{2} + 5 b = 42

b = \frac{7}{2}, b = - 6

Decimal Notation

b = 3.5, b = - 6

Solution steps

2 b^{2} + 5 b = 42

Move

42

to the left side

2 b^{2} + 5 b - 42 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 ( - 42 )}{2 \cdot 2}

5^{2} - 4 \cdot 2 (- 42) = 19

b_{1, 2} = \frac{- 5 \pm 19}{2 \cdot 2}

Separate the solutions

b_{1} = \frac{- 5 + 19}{2 \cdot 2}, b_{2} = \frac{- 5 - 19}{2 \cdot 2}

b = \frac{- 5 + 19}{2 \cdot 2} : \frac{7}{2}

b = \frac{- 5 - 19}{2 \cdot 2} : - 6

The solutions to the quadratic equation are:

b = \frac{7}{2}, b = - 6

y^{2} + 10 y + 9 = 0

t, d = 16 t^{2}

y, x^{2} + 2 x y + y^{2} = 1

9 x^{2} = 54 x - 83

2 x^{2} - 5 x - 3 = 0

What are the solutions to the equation 2b^2+5b=42 ?
The solutions to the equation 2b^2+5b=42 are b= 7/2 ,b=-6
Find the zeros of 2b^2+5b=42
The zeros of 2b^2+5b=42 are b= 7/2 ,b=-6