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-r+3r^2-9=0
We add all the numbers together, and all the variables
3r^2-1r-9=0
a = 3; b = -1; c = -9;
Δ = b2-4ac
Δ = -12-4·3·(-9)
Δ = 109
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{109}}{2*3}=\frac{1-\sqrt{109}}{6} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{109}}{2*3}=\frac{1+\sqrt{109}}{6} $
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