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

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

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

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

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w^{2} - 4 w = 12

w = 6, w = - 2

Solution steps

w^{2} - 4 w = 12

Move

12

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

w = 6, w = - 2

2 x^{2} + 3 x + 1

x^{2} + 6 = 55

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

x, x^{2} y^{3} + 4 x y + x - 6 y = 2

x^{2} + 3 x + 1 = 1

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