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8r^2-18r+9=0
a = 8; b = -18; c = +9;
Δ = b2-4ac
Δ = -182-4·8·9
Δ = 36
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{36}=6$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6}{2*8}=\frac{12}{16} =3/4 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6}{2*8}=\frac{24}{16} =1+1/2 $
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