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96+X^2-48X=0
a = 1; b = -48; c = +96;
Δ = b2-4ac
Δ = -482-4·1·96
Δ = 1920
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1920}=\sqrt{64*30}=\sqrt{64}*\sqrt{30}=8\sqrt{30}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-8\sqrt{30}}{2*1}=\frac{48-8\sqrt{30}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+8\sqrt{30}}{2*1}=\frac{48+8\sqrt{30}}{2} $
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