What are the solutions to the equation 3x^2-136x+1300=0 ?

The solutions to the equation 3x^2-136x+1300=0 are x=(2(34+sqrt(181)))/3 ,x=(2(34-sqrt(181)))/3

Find the zeros of 3x^2-136x+1300=0

The zeros of 3x^2-136x+1300=0 are x=(2(34+sqrt(181)))/3 ,x=(2(34-sqrt(181)))/3

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3 x^{2} - 136 x + 1300 = 0

x = \frac{2 ( 34 + 181 )}{3}, x = \frac{2 ( 34 - 181 )}{3}

Decimal

x = 31.63574 \dots, x = 13.69758 \dots

Solution steps

3 x^{2} - 136 x + 1300 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 136 ) \pm ( - 136 ) ^{2} - 4 \cdot 3 \cdot 1300}{2 \cdot 3}

(- 136)^{2} - 4 \cdot 3 \cdot 1300 = 4181

x_{1, 2} = \frac{- ( - 136 ) \pm 4 181}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 136 ) + 4 181}{2 \cdot 3}, x_{2} = \frac{- ( - 136 ) - 4 181}{2 \cdot 3}

x = \frac{- ( - 136 ) + 4 181}{2 \cdot 3} : \frac{2 ( 34 + 181 )}{3}

x = \frac{- ( - 136 ) - 4 181}{2 \cdot 3} : \frac{2 ( 34 - 181 )}{3}

The solutions to the quadratic equation are:

x = \frac{2 ( 34 + 181 )}{3}, x = \frac{2 ( 34 - 181 )}{3}

4 x^{2} - 6 x + 2 = 5 + 5 x

- 43 = 5 - 3 b^{2}

- 4.9 t^{2} + 12.25 t + 115

x^{2} + 8 = - 17

x, x^{2} + 0^{2} - 4 = 0

What are the solutions to the equation 3x^2-136x+1300=0 ?
The solutions to the equation 3x^2-136x+1300=0 are x=(2(34+sqrt(181)))/3 ,x=(2(34-sqrt(181)))/3
Find the zeros of 3x^2-136x+1300=0
The zeros of 3x^2-136x+1300=0 are x=(2(34+sqrt(181)))/3 ,x=(2(34-sqrt(181)))/3