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

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

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

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

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12 = x^{2} - 4 x

x = 6, x = - 2

Solution steps

12 = x^{2} - 4 x

Switch sides

x^{2} - 4 x = 12

Move

12

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 6, x = - 2

(x - 4)^{2}

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

\frac{1}{81} p - 16 p^{5}

\frac{x ^{- 1} x ^{- 1}}{y ^{6} y ^{- 4}}

3 a (a + b + c) - 2 b (a + b + c) - c (a + b + c)

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