What are the solutions to the equation 3q^2-116q+980=0 ?

The solutions to the equation 3q^2-116q+980=0 are q=(2(29+sqrt(106)))/3 ,q=(2(29-sqrt(106)))/3

Find the zeros of 3q^2-116q+980=0

The zeros of 3q^2-116q+980=0 are q=(2(29+sqrt(106)))/3 ,q=(2(29-sqrt(106)))/3

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3 q^{2} - 116 q + 980 = 0

q = \frac{2 ( 29 + 106 )}{3}, q = \frac{2 ( 29 - 106 )}{3}

Decimal Notation

q = 26.19708 \dots, q = 12.46957 \dots

Solution steps

3 q^{2} - 116 q + 980 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- ( - 116 ) \pm ( - 116 ) ^{2} - 4 \cdot 3 \cdot 980}{2 \cdot 3}

(- 116)^{2} - 4 \cdot 3 \cdot 980 = 4106

q_{1, 2} = \frac{- ( - 116 ) \pm 4 106}{2 \cdot 3}

Separate the solutions

q_{1} = \frac{- ( - 116 ) + 4 106}{2 \cdot 3}, q_{2} = \frac{- ( - 116 ) - 4 106}{2 \cdot 3}

q = \frac{- ( - 116 ) + 4 106}{2 \cdot 3} : \frac{2 ( 29 + 106 )}{3}

q = \frac{- ( - 116 ) - 4 106}{2 \cdot 3} : \frac{2 ( 29 - 106 )}{3}

The solutions to the quadratic equation are:

q = \frac{2 ( 29 + 106 )}{3}, q = \frac{2 ( 29 - 106 )}{3}

- 10 x^{2} - 5 x - 11 = 0

p^{2} + 2 p - 13 = - 3 p

y^{2} - 3 y - 2 = 0

x, p^{1} y = - x^{2} + 5 x - 4

\frac{9}{2} b^{2} = 32

What are the solutions to the equation 3q^2-116q+980=0 ?
The solutions to the equation 3q^2-116q+980=0 are q=(2(29+sqrt(106)))/3 ,q=(2(29-sqrt(106)))/3
Find the zeros of 3q^2-116q+980=0
The zeros of 3q^2-116q+980=0 are q=(2(29+sqrt(106)))/3 ,q=(2(29-sqrt(106)))/3