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(5q-3)(5q+3)=0
We use the square of the difference formula
25q^2-9=0
a = 25; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·25·(-9)
Δ = 900
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}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{900}=30$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-30}{2*25}=\frac{-30}{50} =-3/5 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+30}{2*25}=\frac{30}{50} =3/5 $
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