What are the solutions to the equation 6q^2-13q+6=2q^2 ?

The solutions to the equation 6q^2-13q+6=2q^2 are q=(13+sqrt(73))/8 ,q=(13-sqrt(73))/8

Find the zeros of 6q^2-13q+6=2q^2

The zeros of 6q^2-13q+6=2q^2 are q=(13+sqrt(73))/8 ,q=(13-sqrt(73))/8

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6 q^{2} - 13 q + 6 = 2 q^{2}

q = \frac{13 + 73}{8}, q = \frac{13 - 73}{8}

Decimal Notation

q = 2.69300 \dots, q = 0.55699 \dots

Solution steps

6 q^{2} - 13 q + 6 = 2 q^{2}

Move

2 q^{2}

to the left side

4 q^{2} - 13 q + 6 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 4 \cdot 6}{2 \cdot 4}

(- 13)^{2} - 4 \cdot 4 \cdot 6 = 73

q_{1, 2} = \frac{- ( - 13 ) \pm 73}{2 \cdot 4}

Separate the solutions

q_{1} = \frac{- ( - 13 ) + 73}{2 \cdot 4}, q_{2} = \frac{- ( - 13 ) - 73}{2 \cdot 4}

q = \frac{- ( - 13 ) + 73}{2 \cdot 4} : \frac{13 + 73}{8}

q = \frac{- ( - 13 ) - 73}{2 \cdot 4} : \frac{13 - 73}{8}

The solutions to the quadratic equation are:

q = \frac{13 + 73}{8}, q = \frac{13 - 73}{8}

x^{2} + 5 x - 6 = 0

u \cdot u^{\frac{1}{2}}

2704

\frac{1}{a ^{5} a ^{- 4}}

4 x - 28 = 5 x \cdot (x - 7)

What are the solutions to the equation 6q^2-13q+6=2q^2 ?
The solutions to the equation 6q^2-13q+6=2q^2 are q=(13+sqrt(73))/8 ,q=(13-sqrt(73))/8
Find the zeros of 6q^2-13q+6=2q^2
The zeros of 6q^2-13q+6=2q^2 are q=(13+sqrt(73))/8 ,q=(13-sqrt(73))/8