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2+A2=96
We move all terms to the left:
2+A2-(96)=0
We add all the numbers together, and all the variables
A^2-94=0
a = 1; b = 0; c = -94;
Δ = b2-4ac
Δ = 02-4·1·(-94)
Δ = 376
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{376}=\sqrt{4*94}=\sqrt{4}*\sqrt{94}=2\sqrt{94}$$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{94}}{2*1}=\frac{0-2\sqrt{94}}{2} =-\frac{2\sqrt{94}}{2} =-\sqrt{94} $$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{94}}{2*1}=\frac{0+2\sqrt{94}}{2} =\frac{2\sqrt{94}}{2} =\sqrt{94} $
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