What is the inverse of 0=4x^2-28x+16y-31 ?

The inverse of 0=4x^2-28x+16y-31 is -(-7+4sqrt(-x+5))/2 ,(4sqrt(-x+5)+7)/2

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inverse of 0=4x^2-28x+16y-31

inverse

0 = 4 x^{2} - 28 x + 16 y - 31

- \frac{- 7 + 4 - x + 5}{2}, \frac{4 - x + 5 + 7}{2}

Solution steps

y = \frac{- 4 x ^{2} + 28 x + 31}{16}

Replace

x

with

y

x = \frac{- 4 y ^{2} + 28 y + 31}{16}

Solve for

y, x = \frac{- 4 y ^{2} + 28 y + 31}{16}

- \frac{- 7 + 4 - x + 5}{2}, \frac{4 - x + 5 + 7}{2}

f (x) = - x^{2} + 4 x

y = - x + 1

f (x) = \frac{5}{3 x + 2}

f (x) = 0.25 e^{- 3 x}

f (x) = lo g_{2} (x) - lo g_{2} (12)

What is the inverse of 0=4x^2-28x+16y-31 ?
The inverse of 0=4x^2-28x+16y-31 is -(-7+4sqrt(-x+5))/2 ,(4sqrt(-x+5)+7)/2