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12^2+x^2=20^2
We move all terms to the left:
12^2+x^2-(20^2)=0
We add all the numbers together, and all the variables
x^2-256=0
a = 1; b = 0; c = -256;
Δ = b2-4ac
Δ = 02-4·1·(-256)
Δ = 1024
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{1024}=32$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32}{2*1}=\frac{-32}{2} =-16 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32}{2*1}=\frac{32}{2} =16 $
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