What are the solutions to the equation -y^2-4y+12=0 ?

The solutions to the equation -y^2-4y+12=0 are y=-6,y=2

Find the zeros of -y^2-4y+12=0

The zeros of -y^2-4y+12=0 are y=-6,y=2

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

y = - 6, y = 2

Solution steps

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 (- 1) \cdot 12 = 8

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

Separate the solutions

y_{1} = \frac{- ( - 4 ) + 8}{2 ( - 1 )}, y_{2} = \frac{- ( - 4 ) - 8}{2 ( - 1 )}

y = \frac{- ( - 4 ) + 8}{2 ( - 1 )} : - 6

y = \frac{- ( - 4 ) - 8}{2 ( - 1 )} : 2

The solutions to the quadratic equation are:

y = - 6, y = 2

\frac{5}{12} + \frac{1}{3} - \frac{2}{5}

(3 x - 4)^{2} = 42

x^{2} - 60 x + 851

- x^{2} + 25 = 0

\frac{0.8}{2 ^{6}}

What are the solutions to the equation -y^2-4y+12=0 ?
The solutions to the equation -y^2-4y+12=0 are y=-6,y=2
Find the zeros of -y^2-4y+12=0
The zeros of -y^2-4y+12=0 are y=-6,y=2