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14-2c^2-3c=0
a = -2; b = -3; c = +14;
Δ = b2-4ac
Δ = -32-4·(-2)·14
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$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-11}{2*-2}=\frac{-8}{-4} =+2 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*-2}=\frac{14}{-4} =-3+1/2 $
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