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12m^2+28m-40=0
a = 12; b = 28; c = -40;
Δ = b2-4ac
Δ = 282-4·12·(-40)
Δ = 2704
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{2704}=52$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-52}{2*12}=\frac{-80}{24} =-3+1/3 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+52}{2*12}=\frac{24}{24} =1 $
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