What are the solutions to the equation y^2+7y-8=0 ?

The solutions to the equation y^2+7y-8=0 are y=1,y=-8

Find the zeros of y^2+7y-8=0

The zeros of y^2+7y-8=0 are y=1,y=-8

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y^{2} + 7 y - 8 = 0

y = 1, y = - 8

Solution steps

y^{2} + 7 y - 8 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot (- 8) = 9

y_{1, 2} = \frac{- 7 \pm 9}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 7 + 9}{2 \cdot 1}, y_{2} = \frac{- 7 - 9}{2 \cdot 1}

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

y = \frac{- 7 - 9}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

y = 1, y = - 8

\frac{1}{9} = x^{2}

w^{2} + 4 w - 12 = 0

k^{2} + 3 k = 0

- x^{2} - 100 = 0

49 k^{2} + 7 k - 11 = 1 - 5 k + 4 k^{2}

What are the solutions to the equation y^2+7y-8=0 ?
The solutions to the equation y^2+7y-8=0 are y=1,y=-8
Find the zeros of y^2+7y-8=0
The zeros of y^2+7y-8=0 are y=1,y=-8