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15m(m+1)=m-3
We move all terms to the left:
15m(m+1)-(m-3)=0
We multiply parentheses
15m^2+15m-(m-3)=0
We get rid of parentheses
15m^2+15m-m+3=0
We add all the numbers together, and all the variables
15m^2+14m+3=0
a = 15; b = 14; c = +3;
Δ = b2-4ac
Δ = 142-4·15·3
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$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4}{2*15}=\frac{-18}{30} =-3/5 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4}{2*15}=\frac{-10}{30} =-1/3 $
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