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12m^2+19m+5=0
a = 12; b = 19; c = +5;
Δ = b2-4ac
Δ = 192-4·12·5
Δ = 121
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{121}=11$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-11}{2*12}=\frac{-30}{24} =-1+1/4 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+11}{2*12}=\frac{-8}{24} =-1/3 $
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