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(16x)^2+(9x)^2=52^2
We move all terms to the left:
(16x)^2+(9x)^2-(52^2)=0
We add all the numbers together, and all the variables
25x^2-2704=0
a = 25; b = 0; c = -2704;
Δ = b2-4ac
Δ = 02-4·25·(-2704)
Δ = 270400
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{270400}=520$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-520}{2*25}=\frac{-520}{50} =-10+2/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+520}{2*25}=\frac{520}{50} =10+2/5 $
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