What are the solutions to the equation y^2+18y-40=0 ?

The solutions to the equation y^2+18y-40=0 are y=2,y=-20

Find the zeros of y^2+18y-40=0

The zeros of y^2+18y-40=0 are y=2,y=-20

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y^{2} + 18 y - 40 = 0

y = 2, y = - 20

Solution steps

y^{2} + 18 y - 40 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 1 \cdot ( - 40 )}{2 \cdot 1}

1 8^{2} - 4 \cdot 1 \cdot (- 40) = 22

y_{1, 2} = \frac{- 18 \pm 22}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 18 + 22}{2 \cdot 1}, y_{2} = \frac{- 18 - 22}{2 \cdot 1}

y = \frac{- 18 + 22}{2 \cdot 1} : 2

y = \frac{- 18 - 22}{2 \cdot 1} : - 20

The solutions to the quadratic equation are:

y = 2, y = - 20

x^{2} + 6 x = 55

k (k - 1) = 12

y^{2} + 2 y - 48 = 0

(x + 4) (x - 6) = 5 (x - 6)

\frac{2 x ^{2}}{4} = \frac{x}{15}

What are the solutions to the equation y^2+18y-40=0 ?
The solutions to the equation y^2+18y-40=0 are y=2,y=-20
Find the zeros of y^2+18y-40=0
The zeros of y^2+18y-40=0 are y=2,y=-20