What are the solutions to the equation y^2+41y-371=0 ?

The solutions to the equation y^2+41y-371=0 are y=(-41+sqrt(3165))/2 ,y=(-41-sqrt(3165))/2

Find the zeros of y^2+41y-371=0

The zeros of y^2+41y-371=0 are y=(-41+sqrt(3165))/2 ,y=(-41-sqrt(3165))/2

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y^{2} + 41 y - 371 = 0

y = \frac{- 41 + 3165}{2}, y = \frac{- 41 - 3165}{2}

Decimal Notation

y = 7.62916 \dots, y = - 48.62916 \dots

Solution steps

y^{2} + 41 y - 371 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 41 \pm 4 1 ^{2} - 4 \cdot 1 \cdot ( - 371 )}{2 \cdot 1}

4 1^{2} - 4 \cdot 1 \cdot (- 371) = 3165

y_{1, 2} = \frac{- 41 \pm 3165}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 41 + 3165}{2 \cdot 1}, y_{2} = \frac{- 41 - 3165}{2 \cdot 1}

y = \frac{- 41 + 3165}{2 \cdot 1} : \frac{- 41 + 3165}{2}

y = \frac{- 41 - 3165}{2 \cdot 1} : \frac{- 41 - 3165}{2}

The solutions to the quadratic equation are:

y = \frac{- 41 + 3165}{2}, y = \frac{- 41 - 3165}{2}

3 x^{2} + 5 x - 2

- t^{2} - 1 = 0

1 = 2 π \cdot r^{2}

20 x^{2} = 10 x

d^{2} + (- 5) = - 2

What are the solutions to the equation y^2+41y-371=0 ?
The solutions to the equation y^2+41y-371=0 are y=(-41+sqrt(3165))/2 ,y=(-41-sqrt(3165))/2
Find the zeros of y^2+41y-371=0
The zeros of y^2+41y-371=0 are y=(-41+sqrt(3165))/2 ,y=(-41-sqrt(3165))/2