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(3m-5)(m-3)=0
We multiply parentheses ..
(+3m^2-9m-5m+15)=0
We get rid of parentheses
3m^2-9m-5m+15=0
We add all the numbers together, and all the variables
3m^2-14m+15=0
a = 3; b = -14; c = +15;
Δ = b2-4ac
Δ = -142-4·3·15
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-4}{2*3}=\frac{10}{6} =1+2/3 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+4}{2*3}=\frac{18}{6} =3 $
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