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9b^2+66b+12=0
a = 9; b = 66; c = +12;
Δ = b2-4ac
Δ = 662-4·9·12
Δ = 3924
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3924}=\sqrt{36*109}=\sqrt{36}*\sqrt{109}=6\sqrt{109}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(66)-6\sqrt{109}}{2*9}=\frac{-66-6\sqrt{109}}{18} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(66)+6\sqrt{109}}{2*9}=\frac{-66+6\sqrt{109}}{18} $
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