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3x^2-22x+12=0
a = 3; b = -22; c = +12;
Δ = b2-4ac
Δ = -222-4·3·12
Δ = 340
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{340}=\sqrt{4*85}=\sqrt{4}*\sqrt{85}=2\sqrt{85}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{85}}{2*3}=\frac{22-2\sqrt{85}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{85}}{2*3}=\frac{22+2\sqrt{85}}{6} $
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