19+3x^2=100+2x^2

Solution for 19+3x^2=100+2x^2 equation:

19+3x^2=100+2x^2

We move all terms to the left:

19+3x^2-(100+2x^2)=0

We get rid of parentheses

3x^2-2x^2-100+19=0

We add all the numbers together, and all the variables

x^2-81=0

a = 1; b = 0; c = -81;
Δ = b2-4ac
Δ = 02-4·1·(-81)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18}{2*1}=\frac{-18}{2}
=-9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18}{2*1}=\frac{18}{2}
=9
$