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14r^2-29r+12=0
a = 14; b = -29; c = +12;
Δ = b2-4ac
Δ = -292-4·14·12
Δ = 169
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{169}=13$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-13}{2*14}=\frac{16}{28} =4/7 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+13}{2*14}=\frac{42}{28} =1+1/2 $
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