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2x^2-4x=6
We move all terms to the left:
2x^2-4x-(6)=0
a = 2; b = -4; c = -6;
Δ = b2-4ac
Δ = -42-4·2·(-6)
Δ = 64
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*2}=\frac{-4}{4} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*2}=\frac{12}{4} =3 $
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