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3c^2+13c-30=0
a = 3; b = 13; c = -30;
Δ = b2-4ac
Δ = 132-4·3·(-30)
Δ = 529
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{529}=23$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-23}{2*3}=\frac{-36}{6} =-6 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+23}{2*3}=\frac{10}{6} =1+2/3 $
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