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3c^2+2c-34=0
a = 3; b = 2; c = -34;
Δ = b2-4ac
Δ = 22-4·3·(-34)
Δ = 412
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}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{412}=\sqrt{4*103}=\sqrt{4}*\sqrt{103}=2\sqrt{103}$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{103}}{2*3}=\frac{-2-2\sqrt{103}}{6} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{103}}{2*3}=\frac{-2+2\sqrt{103}}{6} $
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