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

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

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

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

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

x = 8, x = 4

Solution steps

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

Move

x^{2}

to the left side

x^{2} - 12 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 32 = 4

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

Separate the solutions

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

x = \frac{- ( - 12 ) + 4}{2 \cdot 1} : 8

x = \frac{- ( - 12 ) - 4}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

x = 8, x = 4

4 - (- 2)^{2}

\frac{1}{12 m ^{5} n ^{3}} + \frac{6}{18 mn}

\frac{1}{3} (4)

lo g_{7} (49) = x

- 4 ∣ x ∣ = 0

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