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6z^2+11z-35=0
a = 6; b = 11; c = -35;
Δ = b2-4ac
Δ = 112-4·6·(-35)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{961}=31$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-31}{2*6}=\frac{-42}{12} =-3+1/2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+31}{2*6}=\frac{20}{12} =1+2/3 $
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