4x^2+x^2=320

Solution for 4x^2+x^2=320 equation:

4x^2+x^2=320

We move all terms to the left:

4x^2+x^2-(320)=0

We add all the numbers together, and all the variables

5x^2-320=0

a = 5; b = 0; c = -320;
Δ = b2-4ac
Δ = 02-4·5·(-320)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80}{2*5}=\frac{-80}{10}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80}{2*5}=\frac{80}{10}
=8
$