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10z^2-4z=142
We move all terms to the left:
10z^2-4z-(142)=0
a = 10; b = -4; c = -142;
Δ = b2-4ac
Δ = -42-4·10·(-142)
Δ = 5696
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5696}=\sqrt{64*89}=\sqrt{64}*\sqrt{89}=8\sqrt{89}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8\sqrt{89}}{2*10}=\frac{4-8\sqrt{89}}{20} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8\sqrt{89}}{2*10}=\frac{4+8\sqrt{89}}{20} $
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