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f^2-22=0
a = 1; b = 0; c = -22;
Δ = b2-4ac
Δ = 02-4·1·(-22)
Δ = 88
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{22}}{2*1}=\frac{0-2\sqrt{22}}{2} =-\frac{2\sqrt{22}}{2} =-\sqrt{22} $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{22}}{2*1}=\frac{0+2\sqrt{22}}{2} =\frac{2\sqrt{22}}{2} =\sqrt{22} $
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