What are the solutions to the equation 12x^2-72x+96=0 ?

The solutions to the equation 12x^2-72x+96=0 are x=4,x=2

Find the zeros of 12x^2-72x+96=0

The zeros of 12x^2-72x+96=0 are x=4,x=2

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12 x^{2} - 72 x + 96 = 0

x = 4, x = 2

Solution steps

12 x^{2} - 72 x + 96 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 72 ) \pm ( - 72 ) ^{2} - 4 \cdot 12 \cdot 96}{2 \cdot 12}

(- 72)^{2} - 4 \cdot 12 \cdot 96 = 24

x_{1, 2} = \frac{- ( - 72 ) \pm 24}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 72 ) + 24}{2 \cdot 12}, x_{2} = \frac{- ( - 72 ) - 24}{2 \cdot 12}

x = \frac{- ( - 72 ) + 24}{2 \cdot 12} : 4

x = \frac{- ( - 72 ) - 24}{2 \cdot 12} : 2

The solutions to the quadratic equation are:

x = 4, x = 2

- 3 ∣ x - 6 ∣ \geq - 15

2 x^{2} + 4 x - 9 = 2 (x - 1)

\frac{x - 10}{8} = 10

i^{- 2}

What are the solutions to the equation 12x^2-72x+96=0 ?
The solutions to the equation 12x^2-72x+96=0 are x=4,x=2
Find the zeros of 12x^2-72x+96=0
The zeros of 12x^2-72x+96=0 are x=4,x=2