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3x^2+-6x+12=0
We add all the numbers together, and all the variables
3x^2-6x=0
a = 3; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·3·0
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*3}=\frac{0}{6} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*3}=\frac{12}{6} =2 $
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