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

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

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

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

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

x = 2, x = - 6

Solution steps

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

Write in the standard form

a x^{2} + b x + c = 0

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} : 2

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

The solutions to the quadratic equation are:

x = 2, x = - 6

1.5 x^{2} - 45 x - 45 = 0

x, (x - 1)^{2} + y^{2} = z

x, s (x) = (- 2 x^{2} - 7 x + 4) - (x^{2} - 2 x - 1)

x^{2} + 2 x + n^{2} = 0

- 16 t^{2} + 32 t + 384 = 0

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