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3m^2-22m+24=0
a = 3; b = -22; c = +24;
Δ = b2-4ac
Δ = -222-4·3·24
Δ = 196
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{196}=14$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-14}{2*3}=\frac{8}{6} =1+1/3 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+14}{2*3}=\frac{36}{6} =6 $
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