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6c^2+11c-10=0
a = 6; b = 11; c = -10;
Δ = b2-4ac
Δ = 112-4·6·(-10)
Δ = 361
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{361}=19$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-19}{2*6}=\frac{-30}{12} =-2+1/2 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+19}{2*6}=\frac{8}{12} =2/3 $
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