256x^2+81x^2=40

Solution for 256x^2+81x^2=40 equation:

256x^2+81x^2=40

We move all terms to the left:

256x^2+81x^2-(40)=0

We add all the numbers together, and all the variables

337x^2-40=0

a = 337; b = 0; c = -40;
Δ = b2-4ac
Δ = 02-4·337·(-40)
Δ = 53920
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:

$\sqrt{\Delta}=\sqrt{53920}=\sqrt{16*3370}=\sqrt{16}*\sqrt{3370}=4\sqrt{3370}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3370}}{2*337}=\frac{0-4\sqrt{3370}}{674}
=-\frac{4\sqrt{3370}}{674}
=-\frac{2\sqrt{3370}}{337}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3370}}{2*337}=\frac{0+4\sqrt{3370}}{674}
=\frac{4\sqrt{3370}}{674}
=\frac{2\sqrt{3370}}{337}
$