What are the solutions to the equation 39y^2-148y+132=0 ?

The solutions to the equation 39y^2-148y+132=0 are y=(2(37+sqrt(82)))/(39),y=(2(37-sqrt(82)))/(39)

Find the zeros of 39y^2-148y+132=0

The zeros of 39y^2-148y+132=0 are y=(2(37+sqrt(82)))/(39),y=(2(37-sqrt(82)))/(39)

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39 y^{2} - 148 y + 132 = 0

y = \frac{2 ( 37 + 82 )}{39}, y = \frac{2 ( 37 - 82 )}{39}

Decimal Notation

y = 2.36181 \dots, y = 1.43305 \dots

Solution steps

39 y^{2} - 148 y + 132 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 148 ) \pm ( - 148 ) ^{2} - 4 \cdot 39 \cdot 132}{2 \cdot 39}

(- 148)^{2} - 4 \cdot 39 \cdot 132 = 482

y_{1, 2} = \frac{- ( - 148 ) \pm 4 82}{2 \cdot 39}

Separate the solutions

y_{1} = \frac{- ( - 148 ) + 4 82}{2 \cdot 39}, y_{2} = \frac{- ( - 148 ) - 4 82}{2 \cdot 39}

y = \frac{- ( - 148 ) + 4 82}{2 \cdot 39} : \frac{2 ( 37 + 82 )}{39}

y = \frac{- ( - 148 ) - 4 82}{2 \cdot 39} : \frac{2 ( 37 - 82 )}{39}

The solutions to the quadratic equation are:

y = \frac{2 ( 37 + 82 )}{39}, y = \frac{2 ( 37 - 82 )}{39}

(4 x + 1)^{2} + 9 (4 x + 1) = - 18

- 5 x^{2} - 90 x = - 555

x, 5 x^{2} - 12 y^{2} + 3 x y^{2} = 10

x^{2} = - 15

(3 p + 2)^{2} = 8

What are the solutions to the equation 39y^2-148y+132=0 ?
The solutions to the equation 39y^2-148y+132=0 are y=(2(37+sqrt(82)))/(39),y=(2(37-sqrt(82)))/(39)
Find the zeros of 39y^2-148y+132=0
The zeros of 39y^2-148y+132=0 are y=(2(37+sqrt(82)))/(39),y=(2(37-sqrt(82)))/(39)