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3m^2+4m-7=0
a = 3; b = 4; c = -7;
Δ = b2-4ac
Δ = 42-4·3·(-7)
Δ = 100
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{100}=10$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-10}{2*3}=\frac{-14}{6} =-2+1/3 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+10}{2*3}=\frac{6}{6} =1 $
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