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12c^2-28c+15=0
a = 12; b = -28; c = +15;
Δ = b2-4ac
Δ = -282-4·12·15
Δ = 64
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{64}=8$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-8}{2*12}=\frac{20}{24} =5/6 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+8}{2*12}=\frac{36}{24} =1+1/2 $
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