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0=3m^2-16m+21
We move all terms to the left:
0-(3m^2-16m+21)=0
We add all the numbers together, and all the variables
-(3m^2-16m+21)=0
We get rid of parentheses
-3m^2+16m-21=0
a = -3; b = 16; c = -21;
Δ = b2-4ac
Δ = 162-4·(-3)·(-21)
Δ = 4
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{4}=2$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2}{2*-3}=\frac{-18}{-6} =+3 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2}{2*-3}=\frac{-14}{-6} =2+1/3 $
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