What are the solutions to the equation 64x^2-212x+49=0 ?

The solutions to the equation 64x^2-212x+49=0 are x= 49/16 ,x= 1/4

Find the zeros of 64x^2-212x+49=0

The zeros of 64x^2-212x+49=0 are x= 49/16 ,x= 1/4

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64 x^{2} - 212 x + 49 = 0

x = \frac{49}{16}, x = \frac{1}{4}

Decimal Notation

x = 3.0625, x = 0.25

Solution steps

64 x^{2} - 212 x + 49 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 212 ) \pm ( - 212 ) ^{2} - 4 \cdot 64 \cdot 49}{2 \cdot 64}

(- 212)^{2} - 4 \cdot 64 \cdot 49 = 180

x_{1, 2} = \frac{- ( - 212 ) \pm 180}{2 \cdot 64}

Separate the solutions

x_{1} = \frac{- ( - 212 ) + 180}{2 \cdot 64}, x_{2} = \frac{- ( - 212 ) - 180}{2 \cdot 64}

x = \frac{- ( - 212 ) + 180}{2 \cdot 64} : \frac{49}{16}

x = \frac{- ( - 212 ) - 180}{2 \cdot 64} : \frac{1}{4}

The solutions to the quadratic equation are:

x = \frac{49}{16}, x = \frac{1}{4}

x^{2} = 4 x - 29

b^{2} + b - 1 = 0

0 = 12 x^{2} - 12 x

2 = (\frac{1}{2}) (x - 4)^{2}

(x + 1) (x - 3) = 3

What are the solutions to the equation 64x^2-212x+49=0 ?
The solutions to the equation 64x^2-212x+49=0 are x= 49/16 ,x= 1/4
Find the zeros of 64x^2-212x+49=0
The zeros of 64x^2-212x+49=0 are x= 49/16 ,x= 1/4