sqrt(2x^2-3x-5)<sqrt(x-1) inequation

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How to use it?
Inequation:
xy>-4
(log(x^2+9)/log(3))*log((4/5))*(3*x)/(x+2)-log((4/5))*(6-x)<=0
(2^2-x)>2c-3
1-x>3-2*x
Graphing y =:
sqrt(x-1)
Derivative of:
sqrt(x-1)
Integral of d{x}:
sqrt(x-1)
Identical expressions
sqrt(two x^2-3x- five)<sqrt(x- one)
square root of (2x squared minus 3x minus 5) less than square root of (x minus 1)
square root of (two x squared minus 3x minus five) less than square root of (x minus one)
√(2x^2-3x-5)<√(x-1)
sqrt(2x2-3x-5)<sqrt(x-1)
sqrt2x2-3x-5<sqrtx-1
sqrt(2x²-3x-5)<sqrt(x-1)
sqrt(2x to the power of 2-3x-5)<sqrt(x-1)
sqrt2x^2-3x-5<sqrtx-1
Similar expressions
sqrt(2x^2-3x+5)<sqrt(x-1)
sqrt(2x^2-3x-5)<sqrt(x+1)
3^(sqrt(x))+3^(sqrt(x)-1)-3^(sqrt(x)-2)<11
sqrt(2x^2+3x-5)<sqrt(x-1)

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\/  2*x  - 3*x - 5  < \/ x - 1

$$\sqrt{\left(2 x^{2} - 3 x\right) - 5} < \sqrt{x - 1}$$

sqrt(2*x^2 - 3*x - 5) < sqrt(x - 1)

Solving inequality on a graph

Rapid solution 2 [src]

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[5/2, 1 + \/ 3 )

$$x\ in\ \left[\frac{5}{2}, 1 + \sqrt{3}\right)$$

x in Interval.Ropen(5/2, 1 + sqrt(3))

Rapid solution [src]

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And\5/2 <= x, x < 1 + \/ 3 /

$$\frac{5}{2} \leq x \wedge x < 1 + \sqrt{3}$$

(5/2 <= x)∧(x < 1 + sqrt(3))