x2+(3x)2=160

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Solution for x2+(3x)2=160 equation:



x2+(3x)2=160
We move all terms to the left:
x2+(3x)2-(160)=0
We add all the numbers together, and all the variables
4x^2-160=0
a = 4; b = 0; c = -160;
Δ = b2-4ac
Δ = 02-4·4·(-160)
Δ = 2560
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:
$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{10}}{2*4}=\frac{0-16\sqrt{10}}{8} =-\frac{16\sqrt{10}}{8} =-2\sqrt{10} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{10}}{2*4}=\frac{0+16\sqrt{10}}{8} =\frac{16\sqrt{10}}{8} =2\sqrt{10} $

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