What are the solutions to the equation 3x^2-28x+48=0 ?

The solutions to the equation 3x^2-28x+48=0 are x=(2(7+sqrt(13)))/3 ,x=(2(7-sqrt(13)))/3

Find the zeros of 3x^2-28x+48=0

The zeros of 3x^2-28x+48=0 are x=(2(7+sqrt(13)))/3 ,x=(2(7-sqrt(13)))/3

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3 x^{2} - 28 x + 48 = 0

x = \frac{2 ( 7 + 13 )}{3}, x = \frac{2 ( 7 - 13 )}{3}

Decimal Notation

x = 7.07036 \dots, x = 2.26296 \dots

Solution steps

3 x^{2} - 28 x + 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 28 ) \pm ( - 28 ) ^{2} - 4 \cdot 3 \cdot 48}{2 \cdot 3}

(- 28)^{2} - 4 \cdot 3 \cdot 48 = 413

x_{1, 2} = \frac{- ( - 28 ) \pm 4 13}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 28 ) + 4 13}{2 \cdot 3}, x_{2} = \frac{- ( - 28 ) - 4 13}{2 \cdot 3}

x = \frac{- ( - 28 ) + 4 13}{2 \cdot 3} : \frac{2 ( 7 + 13 )}{3}

x = \frac{- ( - 28 ) - 4 13}{2 \cdot 3} : \frac{2 ( 7 - 13 )}{3}

The solutions to the quadratic equation are:

x = \frac{2 ( 7 + 13 )}{3}, x = \frac{2 ( 7 - 13 )}{3}

3 = 12 x^{2}

- 2 x^{2} - 4 x - 3

- 2 x^{2} - 4 x + 1

2 x^{2} + 6 x - 15 = 0

x^{2} = (2.4 \cdot 1 0^{7})^{2} - (7.2 \cdot 1 0^{15}) (0.035)

What are the solutions to the equation 3x^2-28x+48=0 ?
The solutions to the equation 3x^2-28x+48=0 are x=(2(7+sqrt(13)))/3 ,x=(2(7-sqrt(13)))/3
Find the zeros of 3x^2-28x+48=0
The zeros of 3x^2-28x+48=0 are x=(2(7+sqrt(13)))/3 ,x=(2(7-sqrt(13)))/3