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2m(m-5)=m-3
We move all terms to the left:
2m(m-5)-(m-3)=0
We multiply parentheses
2m^2-10m-(m-3)=0
We get rid of parentheses
2m^2-10m-m+3=0
We add all the numbers together, and all the variables
2m^2-11m+3=0
a = 2; b = -11; c = +3;
Δ = b2-4ac
Δ = -112-4·2·3
Δ = 97
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{97}}{2*2}=\frac{11-\sqrt{97}}{4} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{97}}{2*2}=\frac{11+\sqrt{97}}{4} $
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