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

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

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

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

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

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

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

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

5 x^{2} - x - 4 = 0

6 x^{2} + 3 = 7 x

x^{2} + 4 x + 3 = 195

3 x^{2} + 4 x - 32 = 0

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