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10m^2+42m+36=0
a = 10; b = 42; c = +36;
Δ = b2-4ac
Δ = 422-4·10·36
Δ = 324
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{324}=18$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-18}{2*10}=\frac{-60}{20} =-3 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+18}{2*10}=\frac{-24}{20} =-1+1/5 $
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