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4m^2+2m-42=0
a = 4; b = 2; c = -42;
Δ = b2-4ac
Δ = 22-4·4·(-42)
Δ = 676
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{676}=26$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-26}{2*4}=\frac{-28}{8} =-3+1/2 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+26}{2*4}=\frac{24}{8} =3 $
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