What are the solutions to the equation y^2+4y-32=0 ?

The solutions to the equation y^2+4y-32=0 are y=4,y=-8

Find the zeros of y^2+4y-32=0

The zeros of y^2+4y-32=0 are y=4,y=-8

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y^{2} + 4 y - 32 = 0

y = 4, y = - 8

Solution steps

y^{2} + 4 y - 32 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 32 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 32) = 12

y_{1, 2} = \frac{- 4 \pm 12}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 4 + 12}{2 \cdot 1}, y_{2} = \frac{- 4 - 12}{2 \cdot 1}

y = \frac{- 4 + 12}{2 \cdot 1} : 4

y = \frac{- 4 - 12}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

y = 4, y = - 8

4 x^{2} + 6 x - 12 = 0

x^{2} + (x + 6)^{2} = 4 8^{2}

s^{2} + 0.2 s + 1 = 0

x^{2} - 5 x - 750 = 0

x^{2} - 7 x + 22 = 0

What are the solutions to the equation y^2+4y-32=0 ?
The solutions to the equation y^2+4y-32=0 are y=4,y=-8
Find the zeros of y^2+4y-32=0
The zeros of y^2+4y-32=0 are y=4,y=-8