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3c^2+4c-5=0
a = 3; b = 4; c = -5;
Δ = b2-4ac
Δ = 42-4·3·(-5)
Δ = 76
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}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{19}}{2*3}=\frac{-4-2\sqrt{19}}{6} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{19}}{2*3}=\frac{-4+2\sqrt{19}}{6} $
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