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r2-6r+3=0
We add all the numbers together, and all the variables
r^2-6r+3=0
a = 1; b = -6; c = +3;
Δ = b2-4ac
Δ = -62-4·1·3
Δ = 24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{6}}{2*1}=\frac{6-2\sqrt{6}}{2} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{6}}{2*1}=\frac{6+2\sqrt{6}}{2} $
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