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5r^2+4=12r
We move all terms to the left:
5r^2+4-(12r)=0
a = 5; b = -12; c = +4;
Δ = b2-4ac
Δ = -122-4·5·4
Δ = 64
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{64}=8$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8}{2*5}=\frac{4}{10} =2/5 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8}{2*5}=\frac{20}{10} =2 $
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