What are the solutions to the equation 12x^2-288x+1152=0 ?

The solutions to the equation 12x^2-288x+1152=0 are x=4(3+sqrt(3)),x=4(3-sqrt(3))

Find the zeros of 12x^2-288x+1152=0

The zeros of 12x^2-288x+1152=0 are x=4(3+sqrt(3)),x=4(3-sqrt(3))

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12 x^{2} - 288 x + 1152 = 0

x = 4 (3 + 3), x = 4 (3 - 3)

Decimal Notation

x = 18.92820 \dots, x = 5.07179 \dots

Solution steps

12 x^{2} - 288 x + 1152 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 288 ) \pm ( - 288 ) ^{2} - 4 \cdot 12 \cdot 1152}{2 \cdot 12}

(- 288)^{2} - 4 \cdot 12 \cdot 1152 = 963

x_{1, 2} = \frac{- ( - 288 ) \pm 96 3}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 288 ) + 96 3}{2 \cdot 12}, x_{2} = \frac{- ( - 288 ) - 96 3}{2 \cdot 12}

x = \frac{- ( - 288 ) + 96 3}{2 \cdot 12} : 4 (3 + 3)

x = \frac{- ( - 288 ) - 96 3}{2 \cdot 12} : 4 (3 - 3)

The solutions to the quadratic equation are:

x = 4 (3 + 3), x = 4 (3 - 3)

(X + 6) (X - 6) = 13

x^{2} + 17 x + 12 = 0

23 x = - 15 x^{2} - 6

x^{2} - 16 x + 40 = 0

324 + x^{2} = 0

What are the solutions to the equation 12x^2-288x+1152=0 ?
The solutions to the equation 12x^2-288x+1152=0 are x=4(3+sqrt(3)),x=4(3-sqrt(3))
Find the zeros of 12x^2-288x+1152=0
The zeros of 12x^2-288x+1152=0 are x=4(3+sqrt(3)),x=4(3-sqrt(3))