What are the solutions to the equation 3x^2-12x+10=0 ?

The solutions to the equation 3x^2-12x+10=0 are x=(6+sqrt(6))/3 ,x=(6-sqrt(6))/3

Find the zeros of 3x^2-12x+10=0

The zeros of 3x^2-12x+10=0 are x=(6+sqrt(6))/3 ,x=(6-sqrt(6))/3

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3 x^{2} - 12 x + 10 = 0

x = \frac{6 + 6}{3}, x = \frac{6 - 6}{3}

Decimal Notation

x = 2.81649 \dots, x = 1.18350 \dots

Solution steps

3 x^{2} - 12 x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 3 \cdot 10}{2 \cdot 3}

(- 12)^{2} - 4 \cdot 3 \cdot 10 = 26

x_{1, 2} = \frac{- ( - 12 ) \pm 2 6}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 2 6}{2 \cdot 3}, x_{2} = \frac{- ( - 12 ) - 2 6}{2 \cdot 3}

x = \frac{- ( - 12 ) + 2 6}{2 \cdot 3} : \frac{6 + 6}{3}

x = \frac{- ( - 12 ) - 2 6}{2 \cdot 3} : \frac{6 - 6}{3}

The solutions to the quadratic equation are:

x = \frac{6 + 6}{3}, x = \frac{6 - 6}{3}

105

(3 x + 4)^{2} + 9 (3 x + 4) - 10 = 0

∣ x + 2 ∣ \leq 5

24 x^{5} y^{2} - 16 x y

- \frac{2 2}{3} x - 3 \leq \frac{4 3}{2}

What are the solutions to the equation 3x^2-12x+10=0 ?
The solutions to the equation 3x^2-12x+10=0 are x=(6+sqrt(6))/3 ,x=(6-sqrt(6))/3
Find the zeros of 3x^2-12x+10=0
The zeros of 3x^2-12x+10=0 are x=(6+sqrt(6))/3 ,x=(6-sqrt(6))/3