5+20=4x^2

Solution for 5+20=4x^2 equation:

5+20=4x^2

We move all terms to the left:

5+20-(4x^2)=0

We add all the numbers together, and all the variables

-4x^2+25=0

a = -4; b = 0; c = +25;
Δ = b2-4ac
Δ = 02-4·(-4)·25
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*-4}=\frac{-20}{-8}
=2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*-4}=\frac{20}{-8}
=-2+1/2
$