3x^2-55.6x^2+1000=0

Solution for 3x^2-55.6x^2+1000=0 equation:

3x^2-55.6x^2+1000=0

We add all the numbers together, and all the variables

-52.6x^2+1000=0

a = -52.6; b = 0; c = +1000;
Δ = b2-4ac
Δ = 02-4·(-52.6)·1000
Δ = 210400
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:

$\sqrt{\Delta}=\sqrt{210400}=\sqrt{400*526}=\sqrt{400}*\sqrt{526}=20\sqrt{526}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{526}}{2*-52.6}=\frac{0-20\sqrt{526}}{-105.2}
=-\frac{20\sqrt{526}}{-105.2}
=-\frac{4\sqrt{526}}{-21.04}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{526}}{2*-52.6}=\frac{0+20\sqrt{526}}{-105.2}
=\frac{20\sqrt{526}}{-105.2}
=\frac{4\sqrt{526}}{-21.04}
$