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7m^2+61m-18=0
a = 7; b = 61; c = -18;
Δ = b2-4ac
Δ = 612-4·7·(-18)
Δ = 4225
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{4225}=65$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(61)-65}{2*7}=\frac{-126}{14} =-9 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(61)+65}{2*7}=\frac{4}{14} =2/7 $
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