What is the inverse of y=7x^2+4x-3,x>= 0 ?

The inverse of y=7x^2+4x-3,x>= 0 is (-2+sqrt(7x+25))/7 ,-(sqrt(7x+25)+2)/7

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inverse of y=7x^2+4x-3,x>= 0

inverse

y = 7 x^{2} + 4 x - 3, x \geq 0

\frac{- 2 + 7 x + 25}{7}, - \frac{7 x + 25 + 2}{7}

Solution steps

y = 7 x^{2} + 4 x - 3

Replace

x

with

y

x = 7 y^{2} + 4 y - 3

Solve for

y, x = 7 y^{2} + 4 y - 3

\frac{- 2 + 7 x + 25}{7}, - \frac{7 x + 25 + 2}{7}

What is the inverse of y=7x^2+4x-3,x>= 0 ?
The inverse of y=7x^2+4x-3,x>= 0 is (-2+sqrt(7x+25))/7 ,-(sqrt(7x+25)+2)/7