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z^2-10z=24
We move all terms to the left:
z^2-10z-(24)=0
a = 1; b = -10; c = -24;
Δ = b2-4ac
Δ = -102-4·1·(-24)
Δ = 196
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{196}=14$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-14}{2*1}=\frac{-4}{2} =-2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+14}{2*1}=\frac{24}{2} =12 $
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