What are the solutions to the equation 12x^2-288x+1260=0 ?

The solutions to the equation 12x^2-288x+1260=0 are x=12+sqrt(39),x=12-sqrt(39)

Find the zeros of 12x^2-288x+1260=0

The zeros of 12x^2-288x+1260=0 are x=12+sqrt(39),x=12-sqrt(39)

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12 x^{2} - 288 x + 1260 = 0

x = 12 + 39, x = 12 - 39

Decimal Notation

x = 18.24499 \dots, x = 5.75500 \dots

Solution steps

12 x^{2} - 288 x + 1260 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 288 ) \pm ( - 288 ) ^{2} - 4 \cdot 12 \cdot 1260}{2 \cdot 12}

(- 288)^{2} - 4 \cdot 12 \cdot 1260 = 2439

x_{1, 2} = \frac{- ( - 288 ) \pm 24 39}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 288 ) + 24 39}{2 \cdot 12}, x_{2} = \frac{- ( - 288 ) - 24 39}{2 \cdot 12}

x = \frac{- ( - 288 ) + 24 39}{2 \cdot 12} : 12 + 39

x = \frac{- ( - 288 ) - 24 39}{2 \cdot 12} : 12 - 39

The solutions to the quadratic equation are:

x = 12 + 39, x = 12 - 39

v, \frac{m v ^{2}}{r} = q v B

\frac{1}{6} z^{2} + \frac{4}{6} z = 2

x^{2} - 8 x + 16 = 0

x^{2} + 3 x + 3 = - 2 x - 3

289 x^{2} + 378 x + 89 = 0

What are the solutions to the equation 12x^2-288x+1260=0 ?
The solutions to the equation 12x^2-288x+1260=0 are x=12+sqrt(39),x=12-sqrt(39)
Find the zeros of 12x^2-288x+1260=0
The zeros of 12x^2-288x+1260=0 are x=12+sqrt(39),x=12-sqrt(39)