What are the solutions to the equation 3x^2-24x+40=0 ?

The solutions to the equation 3x^2-24x+40=0 are x=(2(6+sqrt(6)))/3 ,x=(2(6-sqrt(6)))/3

Find the zeros of 3x^2-24x+40=0

The zeros of 3x^2-24x+40=0 are x=(2(6+sqrt(6)))/3 ,x=(2(6-sqrt(6)))/3

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3 x^{2} - 24 x + 40 = 0

x = \frac{2 ( 6 + 6 )}{3}, x = \frac{2 ( 6 - 6 )}{3}

Decimal Notation

x = 5.63299 \dots, x = 2.36700 \dots

Solution steps

3 x^{2} - 24 x + 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 3 \cdot 40}{2 \cdot 3}

(- 24)^{2} - 4 \cdot 3 \cdot 40 = 46

x_{1, 2} = \frac{- ( - 24 ) \pm 4 6}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 24 ) + 4 6}{2 \cdot 3}, x_{2} = \frac{- ( - 24 ) - 4 6}{2 \cdot 3}

x = \frac{- ( - 24 ) + 4 6}{2 \cdot 3} : \frac{2 ( 6 + 6 )}{3}

x = \frac{- ( - 24 ) - 4 6}{2 \cdot 3} : \frac{2 ( 6 - 6 )}{3}

The solutions to the quadratic equation are:

x = \frac{2 ( 6 + 6 )}{3}, x = \frac{2 ( 6 - 6 )}{3}

x^{2} - 10 = 2 x^{2}

n^{2} + 6 n + 9 = 0

x^{2} - x - 0 = 0

(2 x - 3)^{2} - 6 = 43

k^{2} - 8 k - 16 = 0

What are the solutions to the equation 3x^2-24x+40=0 ?
The solutions to the equation 3x^2-24x+40=0 are x=(2(6+sqrt(6)))/3 ,x=(2(6-sqrt(6)))/3
Find the zeros of 3x^2-24x+40=0
The zeros of 3x^2-24x+40=0 are x=(2(6+sqrt(6)))/3 ,x=(2(6-sqrt(6)))/3