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

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

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

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

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y^{2} + y - 42 = 0

y = 6, y = - 7

Solution steps

y^{2} + y - 42 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 6, y = - 7

(x - 6) (x - 8) = - 1

- 3 x^{2} - 18 x - 26 = 0

0.2 x^{2} - 6 x + 80 = 0.6 x^{2} - 12 x + 80

a, c = ba + a^{2}

- 2 x^{2} + 12 x - 8 = 0

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