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a2-12a+4=0
We add all the numbers together, and all the variables
a^2-12a+4=0
a = 1; b = -12; c = +4;
Δ = b2-4ac
Δ = -122-4·1·4
Δ = 128
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8\sqrt{2}}{2*1}=\frac{12-8\sqrt{2}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8\sqrt{2}}{2*1}=\frac{12+8\sqrt{2}}{2} $
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