3x^2+20x^2-100=0

Solution for 3x^2+20x^2-100=0 equation:

3x^2+20x^2-100=0

We add all the numbers together, and all the variables

23x^2-100=0

a = 23; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·23·(-100)
Δ = 9200
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:

$\sqrt{\Delta}=\sqrt{9200}=\sqrt{400*23}=\sqrt{400}*\sqrt{23}=20\sqrt{23}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{23}}{2*23}=\frac{0-20\sqrt{23}}{46}
=-\frac{20\sqrt{23}}{46}
=-\frac{10\sqrt{23}}{23}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{23}}{2*23}=\frac{0+20\sqrt{23}}{46}
=\frac{20\sqrt{23}}{46}
=\frac{10\sqrt{23}}{23}
$