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2m^2+13m+21=0
a = 2; b = 13; c = +21;
Δ = b2-4ac
Δ = 132-4·2·21
Δ = 1
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{1}=1$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-1}{2*2}=\frac{-14}{4} =-3+1/2 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+1}{2*2}=\frac{-12}{4} =-3 $
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