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9z^2+29=36z
We move all terms to the left:
9z^2+29-(36z)=0
a = 9; b = -36; c = +29;
Δ = b2-4ac
Δ = -362-4·9·29
Δ = 252
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-6\sqrt{7}}{2*9}=\frac{36-6\sqrt{7}}{18} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+6\sqrt{7}}{2*9}=\frac{36+6\sqrt{7}}{18} $
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