(9/41)^2=x

Solution for (9/41)^2=x equation:

(9/41)^2=x

We move all terms to the left:

(9/41)^2-(x)=0

We add all the numbers together, and all the variables

-x+(+9/41)^2=0

We add all the numbers together, and all the variables

-1x+(+9/41)^2=0

We multiply all the terms by the denominator


-1x*41)^2+(+9=0

Wy multiply elements

-41x^2+9=0

a = -41; b = 0; c = +9;
Δ = b2-4ac
Δ = 02-4·(-41)·9
Δ = 1476
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1476}=\sqrt{36*41}=\sqrt{36}*\sqrt{41}=6\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{41}}{2*-41}=\frac{0-6\sqrt{41}}{-82}
=-\frac{6\sqrt{41}}{-82}
=-\frac{3\sqrt{41}}{-41}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{41}}{2*-41}=\frac{0+6\sqrt{41}}{-82}
=\frac{6\sqrt{41}}{-82}
=\frac{3\sqrt{41}}{-41}
$