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(2b-5)(3b-7)=0
We multiply parentheses ..
(+6b^2-14b-15b+35)=0
We get rid of parentheses
6b^2-14b-15b+35=0
We add all the numbers together, and all the variables
6b^2-29b+35=0
a = 6; b = -29; c = +35;
Δ = b2-4ac
Δ = -292-4·6·35
Δ = 1
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{1}=1$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-1}{2*6}=\frac{28}{12} =2+1/3 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+1}{2*6}=\frac{30}{12} =2+1/2 $
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