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6z^2-29z-5=0
a = 6; b = -29; c = -5;
Δ = b2-4ac
Δ = -292-4·6·(-5)
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$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-31}{2*6}=\frac{-2}{12} =-1/6 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+31}{2*6}=\frac{60}{12} =5 $
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