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3x^2-22x+5=0
a = 3; b = -22; c = +5;
Δ = b2-4ac
Δ = -222-4·3·5
Δ = 424
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{424}=\sqrt{4*106}=\sqrt{4}*\sqrt{106}=2\sqrt{106}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{106}}{2*3}=\frac{22-2\sqrt{106}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{106}}{2*3}=\frac{22+2\sqrt{106}}{6} $
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