5x^2+4x^2=19^2

Solution for 5x^2+4x^2=19^2 equation:

5x^2+4x^2=19^2

We move all terms to the left:

5x^2+4x^2-(19^2)=0

We add all the numbers together, and all the variables

9x^2-361=0

a = 9; b = 0; c = -361;
Δ = b2-4ac
Δ = 02-4·9·(-361)
Δ = 12996
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12996}=114$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-114}{2*9}=\frac{-114}{18}
=-6+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+114}{2*9}=\frac{114}{18}
=6+1/3
$