(4x^2)+(41)=81

Solution for (4x^2)+(41)=81 equation:

(4x^2)+(41)=81

We move all terms to the left:

(4x^2)+(41)-(81)=0

We add all the numbers together, and all the variables

4x^2-40=0

a = 4; b = 0; c = -40;
Δ = b2-4ac
Δ = 02-4·4·(-40)
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{10}}{2*4}=\frac{0-8\sqrt{10}}{8}
=-\frac{8\sqrt{10}}{8}
=-\sqrt{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{10}}{2*4}=\frac{0+8\sqrt{10}}{8}
=\frac{8\sqrt{10}}{8}
=\sqrt{10}
$