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2r^2-9r+7=0
a = 2; b = -9; c = +7;
Δ = b2-4ac
Δ = -92-4·2·7
Δ = 25
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{25}=5$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-5}{2*2}=\frac{4}{4} =1 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+5}{2*2}=\frac{14}{4} =3+1/2 $
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