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12c^2-24c+5=0
a = 12; b = -24; c = +5;
Δ = b2-4ac
Δ = -242-4·12·5
Δ = 336
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}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4\sqrt{21}}{2*12}=\frac{24-4\sqrt{21}}{24} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4\sqrt{21}}{2*12}=\frac{24+4\sqrt{21}}{24} $
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