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12c^2+29c+15=0
a = 12; b = 29; c = +15;
Δ = b2-4ac
Δ = 292-4·12·15
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{121}=11$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-11}{2*12}=\frac{-40}{24} =-1+2/3 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+11}{2*12}=\frac{-18}{24} =-3/4 $
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