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8x^2-22x+6=0
a = 8; b = -22; c = +6;
Δ = b2-4ac
Δ = -222-4·8·6
Δ = 292
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{73}}{2*8}=\frac{22-2\sqrt{73}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{73}}{2*8}=\frac{22+2\sqrt{73}}{16} $
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