What are the solutions to the equation 27ay-14y^2=10a^2 ?

The solutions to the equation 27ay-14y^2=10a^2 are y= a/2 ,y=(10a)/7

Find the zeros of 27ay-14y^2=10a^2

The zeros of 27ay-14y^2=10a^2 are y= a/2 ,y=(10a)/7

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27 a y - 14 y^{2} = 10 a^{2}

y = \frac{a}{2}, y = \frac{10 a}{7}

Solution steps

27 a y - 14 y^{2} = 10 a^{2}

Move

10 a^{2}

to the left side

27 a y - 14 y^{2} - 10 a^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 14 y^{2} + 27 a y - 10 a^{2} = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 27 a \pm ( 27 a ) ^{2} - 4 ( - 14 ) ( - 10 a ^{2} )}{2 ( - 14 )}

Simplify

(27 a)^{2} - 4 (- 14) (- 10 a^{2}) : 13 a

y_{1, 2} = \frac{- 27 a \pm 13 a}{2 ( - 14 )}

Separate the solutions

y_{1} = \frac{- 27 a + 13 a}{2 ( - 14 )}, y_{2} = \frac{- 27 a - 13 a}{2 ( - 14 )}

y = \frac{- 27 a + 13 a}{2 ( - 14 )} : \frac{a}{2}

y = \frac{- 27 a - 13 a}{2 ( - 14 )} : \frac{10 a}{7}

The solutions to the quadratic equation are:

y = \frac{a}{2}, y = \frac{10 a}{7}

x^{2} + 11 x + 11 = 4 x

x^{2} - 5 x = 0

x - 36 = x^{2} - 11 x

3 x^{2} + 1 = 5

x^{2} + 5 = 30

What are the solutions to the equation 27ay-14y^2=10a^2 ?
The solutions to the equation 27ay-14y^2=10a^2 are y= a/2 ,y=(10a)/7
Find the zeros of 27ay-14y^2=10a^2
The zeros of 27ay-14y^2=10a^2 are y= a/2 ,y=(10a)/7