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2x^2-6x+4=2
We move all terms to the left:
2x^2-6x+4-(2)=0
We add all the numbers together, and all the variables
2x^2-6x+2=0
a = 2; b = -6; c = +2;
Δ = b2-4ac
Δ = -62-4·2·2
Δ = 20
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{5}}{2*2}=\frac{6-2\sqrt{5}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{5}}{2*2}=\frac{6+2\sqrt{5}}{4} $
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