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20m^2+20m-40=0
a = 20; b = 20; c = -40;
Δ = b2-4ac
Δ = 202-4·20·(-40)
Δ = 3600
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{3600}=60$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-60}{2*20}=\frac{-80}{40} =-2 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+60}{2*20}=\frac{40}{40} =1 $
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