What are the solutions to the equation y^2+13y+42=0 ?

The solutions to the equation y^2+13y+42=0 are y=-6,y=-7

Find the zeros of y^2+13y+42=0

The zeros of y^2+13y+42=0 are y=-6,y=-7

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y^{2} + 13 y + 42 = 0

y = - 6, y = - 7

Solution steps

y^{2} + 13 y + 42 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 1 \cdot 42}{2 \cdot 1}

1 3^{2} - 4 \cdot 1 \cdot 42 = 1

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = - 6, y = - 7

x, 7 x^{2} + 3 x - 2 = 0

(2 (\frac{3}{2}) - 1) x^{2} + 4 x + 2 = 0

(x - 3) (x + 3) = 6 x - 14

2 x^{2} - 25 x = 0

(4 c + 2) (2 c - 7) = 0

What are the solutions to the equation y^2+13y+42=0 ?
The solutions to the equation y^2+13y+42=0 are y=-6,y=-7
Find the zeros of y^2+13y+42=0
The zeros of y^2+13y+42=0 are y=-6,y=-7