20^2=4x^2-(2)(0.25)(40)

Solution for 20^2=4x^2-(2)(0.25)(40) equation:

20^2=4x^2-(2)(0.25)(40)

We move all terms to the left:

20^2-(4x^2-(2)(0.25)(40))=0

We add all the numbers together, and all the variables

-(4x^2-2(0.25)40)+400=0


We calculate terms in parentheses: -(4x^2-2(0.25)40), so:

4x^2-2(0.25)40

We add all the numbers together, and all the variables

4x^2-20

Back to the equation:

-(4x^2-20)


We get rid of parentheses

-4x^2+20+400=0

We add all the numbers together, and all the variables

-4x^2+420=0

a = -4; b = 0; c = +420;
Δ = b2-4ac
Δ = 02-4·(-4)·420
Δ = 6720
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:

$\sqrt{\Delta}=\sqrt{6720}=\sqrt{64*105}=\sqrt{64}*\sqrt{105}=8\sqrt{105}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{105}}{2*-4}=\frac{0-8\sqrt{105}}{-8}
=-\frac{8\sqrt{105}}{-8}
=-\frac{\sqrt{105}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{105}}{2*-4}=\frac{0+8\sqrt{105}}{-8}
=\frac{8\sqrt{105}}{-8}
=\frac{\sqrt{105}}{-1}
$