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(3b+5)(3b-5)=24
We move all terms to the left:
(3b+5)(3b-5)-(24)=0
We use the square of the difference formula
9b^2-25-24=0
We add all the numbers together, and all the variables
9b^2-49=0
a = 9; b = 0; c = -49;
Δ = b2-4ac
Δ = 02-4·9·(-49)
Δ = 1764
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1764}=42$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-42}{2*9}=\frac{-42}{18} =-2+1/3 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+42}{2*9}=\frac{42}{18} =2+1/3 $
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