What are the solutions to the equation y^2+7y-18=0 ?

The solutions to the equation y^2+7y-18=0 are y=2,y=-9

Find the zeros of y^2+7y-18=0

The zeros of y^2+7y-18=0 are y=2,y=-9

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y^{2} + 7 y - 18 = 0

y = 2, y = - 9

Solution steps

y^{2} + 7 y - 18 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot ( - 18 )}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot (- 18) = 11

y_{1, 2} = \frac{- 7 \pm 11}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 7 + 11}{2 \cdot 1}, y_{2} = \frac{- 7 - 11}{2 \cdot 1}

y = \frac{- 7 + 11}{2 \cdot 1} : 2

y = \frac{- 7 - 11}{2 \cdot 1} : - 9

The solutions to the quadratic equation are:

y = 2, y = - 9

z^{2} - z + 5 = 0

v^{2} + 18 v + 81 = 50

4 x^{2} = 32 x - 64

x^{2} - 11 x - 105 = - 7 x^{2}

9 m^{2} + m + 6 = 0

What are the solutions to the equation y^2+7y-18=0 ?
The solutions to the equation y^2+7y-18=0 are y=2,y=-9
Find the zeros of y^2+7y-18=0
The zeros of y^2+7y-18=0 are y=2,y=-9