What is the inverse of f(x)=2x^2-3x ?

The inverse of f(x)=2x^2-3x is (3+sqrt(9+8x))/4 ,(3-sqrt(9+8x))/4

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inverse of f(x)=2x^2-3x

inverse

f (x) = 2 x^{2} - 3 x

\frac{3 + 9 + 8 x}{4}, \frac{3 - 9 + 8 x}{4}

Solution steps

y = 2 x^{2} - 3 x

Replace

x

with

y

x = 2 y^{2} - 3 y

Solve for

y, x = 2 y^{2} - 3 y

\frac{3 + 9 + 8 x}{4}, \frac{3 - 9 + 8 x}{4}

\frac{1}{( x + 1 ) ^{2}}

f (x) = \frac{4 x ^{2} - 12 x}{x ^{2} - 9}

\frac{x}{1 - cos ( x )}

y = - 2 sin (\frac{2 π}{7} x)

h (t) = 160 t - 16 t^{2}

What is the inverse of f(x)=2x^2-3x ?
The inverse of f(x)=2x^2-3x is (3+sqrt(9+8x))/4 ,(3-sqrt(9+8x))/4