2x^2+49,2x^2-20=0

Solution for 2x^2+49,2x^2-20=0 equation:

2x^2+49.2x^2-20=0

We add all the numbers together, and all the variables

51.2x^2-20=0

a = 51.2; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·51.2·(-20)
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-64}{2*51.2}=\frac{-64}{102.4}
=-32/51.2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+64}{2*51.2}=\frac{64}{102.4}
=32/51.2
$