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12r^2-10r=0
a = 12; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·12·0
Δ = 100
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{100}=10$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2*12}=\frac{0}{24} =0 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*12}=\frac{20}{24} =5/6 $
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