Solve $2x^2-5x+2=$ $\frac{5-\sqrt{9+8x}}{4}$ Solve $2x^2-5x+2$= $\frac{5-\sqrt{9+8x}}{4}$
I simply do square both sides solve it and I get two value of x one is 2 and other is $\frac{3-√5}{2}$ but this approach it take more time so is there any approach for solving this equation.
 A: Taking Jerry Chang's observation (which amounts to the fact that the expression on the right-hand side is what you get when you plug the coefficients of the quadratic into the quadratic formula, just choosing the minus sign for the square root) and setting $y=2x^2-5x+2$ so that $x=2y^2-5y+2$ we can subtract one of these from the other to obtain:
$$y-x=2(x^2-y^2)-5(x-y)$$
Which yields $y=x$ ; or
$1=5-2(x+y)$ ie $x+y=2$
Then the problem splits as $2x^2-5x+2=x$ or $2x^2-5x+2=2-x$
The solutions of these equations have to be plugged back into the original for checking to see which belongs to which choice of sign of the square root.
A: I don't know another way then going through the algebra, I got the same solutions.
$$\begin{align} 2x^2-5x+2 &= \frac{5-\sqrt{9+8x}}{4} \\
8x^2-20x+3 &= -\sqrt{9+8x} \\
(8x^2-20x+3)^2 &= (-\sqrt{9+8x})^2 \\
64x^4-320x^3+448x^2-120x+9 &= 9 +8x\\
64x^4-320x^3+440x^2-120x &= 0 \\
64x(x-2)(x^2-3x+1) &=0 \\
\end{align}$$
We can see the solutions to that are $0$, $x=2$, $x = \frac{3 + \sqrt{5}}{2}$, $x = \frac{3 - \sqrt{5}}{2}$. Then plugging those into the original equation we get $x=2$ and $x = \frac{3 - \sqrt{5}}{2}$ 
