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x^2-22x+67=0
a = 1; b = -22; c = +67;
Δ = b2-4ac
Δ = -222-4·1·67
Δ = 216
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-6\sqrt{6}}{2*1}=\frac{22-6\sqrt{6}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+6\sqrt{6}}{2*1}=\frac{22+6\sqrt{6}}{2} $
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