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

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

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

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

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y^{2} - 4 y = 12

y = 6, y = - 2

Solution steps

y^{2} - 4 y = 12

Move

12

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 6, y = - 2

30.7886 X^{2} + 68.4915 X - 60 = 0

x (x - 12) = 38

w^{2} = 3 (w - 2)

x^{2} - 2 = 47

x^{2} = 3^{2}

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