What are the solutions to the equation y^2+13y+40=0 ?

The solutions to the equation y^2+13y+40=0 are y=-5,y=-8

Find the zeros of y^2+13y+40=0

The zeros of y^2+13y+40=0 are y=-5,y=-8

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y^{2} + 13 y + 40 = 0

y = - 5, y = - 8

Solution steps

y^{2} + 13 y + 40 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 1 \cdot 40}{2 \cdot 1}

1 3^{2} - 4 \cdot 1 \cdot 40 = 3

y_{1, 2} = \frac{- 13 \pm 3}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 13 + 3}{2 \cdot 1}, y_{2} = \frac{- 13 - 3}{2 \cdot 1}

y = \frac{- 13 + 3}{2 \cdot 1} : - 5

y = \frac{- 13 - 3}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

y = - 5, y = - 8

300 = t^{2} + 2 t

- x^{2} + 21 x - 74 = 0

x, 2 y + 6 x = 3 x^{2} - y^{4}

- 2 x^{2} - 3 x + 20 = 0

x^{2} - 2 (x - 6) = 4 x + 4

What are the solutions to the equation y^2+13y+40=0 ?
The solutions to the equation y^2+13y+40=0 are y=-5,y=-8
Find the zeros of y^2+13y+40=0
The zeros of y^2+13y+40=0 are y=-5,y=-8