What are the solutions to the equation x^2-8x+32=12x-4 ?

The solutions to the equation x^2-8x+32=12x-4 are x=18,x=2

Find the zeros of x^2-8x+32=12x-4

The zeros of x^2-8x+32=12x-4 are x=18,x=2

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x^{2} - 8 x + 32 = 12 x - 4

x = 18, x = 2

Solution steps

x^{2} - 8 x + 32 = 12 x - 4

Move

4

to the left side

x^{2} - 8 x + 36 = 12 x

Move

12 x

to the left side

x^{2} - 20 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

(- 20)^{2} - 4 \cdot 1 \cdot 36 = 16

x_{1, 2} = \frac{- ( - 20 ) \pm 16}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 16}{2 \cdot 1}, x_{2} = \frac{- ( - 20 ) - 16}{2 \cdot 1}

x = \frac{- ( - 20 ) + 16}{2 \cdot 1} : 18

x = \frac{- ( - 20 ) - 16}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

x = 18, x = 2

- 5 x^{2} + 525 x - 2500 = 0

27 = 3 x^{2} + 12 x

x^{2} + 20 = 24

x, y = (x - 3)^{2} - 4

x, y^{4} + 7 x y^{2} = x^{2}

What are the solutions to the equation x^2-8x+32=12x-4 ?
The solutions to the equation x^2-8x+32=12x-4 are x=18,x=2
Find the zeros of x^2-8x+32=12x-4
The zeros of x^2-8x+32=12x-4 are x=18,x=2