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-5=-9q^2
We move all terms to the left:
-5-(-9q^2)=0
We get rid of parentheses
9q^2-5=0
a = 9; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·9·(-5)
Δ = 180
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{5}}{2*9}=\frac{0-6\sqrt{5}}{18} =-\frac{6\sqrt{5}}{18} =-\frac{\sqrt{5}}{3} $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{5}}{2*9}=\frac{0+6\sqrt{5}}{18} =\frac{6\sqrt{5}}{18} =\frac{\sqrt{5}}{3} $
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