What are the solutions to the equation 12x^2-228x+740=0 ?

The solutions to the equation 12x^2-228x+740=0 are x=(57+7sqrt(21))/6 ,x=(57-7sqrt(21))/6

Find the zeros of 12x^2-228x+740=0

The zeros of 12x^2-228x+740=0 are x=(57+7sqrt(21))/6 ,x=(57-7sqrt(21))/6

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12 x^{2} - 228 x + 740 = 0

x = \frac{57 + 7 21}{6}, x = \frac{57 - 7 21}{6}

Decimal Notation

x = 14.84633 \dots, x = 4.15366 \dots

Solution steps

12 x^{2} - 228 x + 740 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 228 ) \pm ( - 228 ) ^{2} - 4 \cdot 12 \cdot 740}{2 \cdot 12}

(- 228)^{2} - 4 \cdot 12 \cdot 740 = 2821

x_{1, 2} = \frac{- ( - 228 ) \pm 28 21}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 228 ) + 28 21}{2 \cdot 12}, x_{2} = \frac{- ( - 228 ) - 28 21}{2 \cdot 12}

x = \frac{- ( - 228 ) + 28 21}{2 \cdot 12} : \frac{57 + 7 21}{6}

x = \frac{- ( - 228 ) - 28 21}{2 \cdot 12} : \frac{57 - 7 21}{6}

The solutions to the quadratic equation are:

x = \frac{57 + 7 21}{6}, x = \frac{57 - 7 21}{6}

(x^{2} - 8 x + 7) = 0

f = (\frac{1}{7})^{1}

(2 y)^{4}

> 0 = x^{2} - 1

9 x^{2} + 54 x = 0

What are the solutions to the equation 12x^2-228x+740=0 ?
The solutions to the equation 12x^2-228x+740=0 are x=(57+7sqrt(21))/6 ,x=(57-7sqrt(21))/6
Find the zeros of 12x^2-228x+740=0
The zeros of 12x^2-228x+740=0 are x=(57+7sqrt(21))/6 ,x=(57-7sqrt(21))/6