What are the solutions to the equation 49x^2-21x+2=0 ?

The solutions to the equation 49x^2-21x+2=0 are x= 2/7 ,x= 1/7

Find the zeros of 49x^2-21x+2=0

The zeros of 49x^2-21x+2=0 are x= 2/7 ,x= 1/7

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49 x^{2} - 21 x + 2 = 0

x = \frac{2}{7}, x = \frac{1}{7}

Solution steps

49 x^{2} - 21 x + 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 21 ) \pm ( - 21 ) ^{2} - 4 \cdot 49 \cdot 2}{2 \cdot 49}

(- 21)^{2} - 4 \cdot 49 \cdot 2 = 7

x_{1, 2} = \frac{- ( - 21 ) \pm 7}{2 \cdot 49}

Separate the solutions

x_{1} = \frac{- ( - 21 ) + 7}{2 \cdot 49}, x_{2} = \frac{- ( - 21 ) - 7}{2 \cdot 49}

x = \frac{- ( - 21 ) + 7}{2 \cdot 49} : \frac{2}{7}

x = \frac{- ( - 21 ) - 7}{2 \cdot 49} : \frac{1}{7}

The solutions to the quadratic equation are:

x = \frac{2}{7}, x = \frac{1}{7}

2 x^{2} = \frac{a ^{3}}{2}

t (6 - t) = 0

175 - 50 x + x^{2} = 0

(3 x - 2)^{2} + 16 = 0

4 (x - 100)^{2} = 16

What are the solutions to the equation 49x^2-21x+2=0 ?
The solutions to the equation 49x^2-21x+2=0 are x= 2/7 ,x= 1/7
Find the zeros of 49x^2-21x+2=0
The zeros of 49x^2-21x+2=0 are x= 2/7 ,x= 1/7