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10z^2+21z-10=0
a = 10; b = 21; c = -10;
Δ = b2-4ac
Δ = 212-4·10·(-10)
Δ = 841
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{841}=29$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-29}{2*10}=\frac{-50}{20} =-2+1/2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+29}{2*10}=\frac{8}{20} =2/5 $
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