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4m^2+6m-4=0
a = 4; b = 6; c = -4;
Δ = b2-4ac
Δ = 62-4·4·(-4)
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$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-10}{2*4}=\frac{-16}{8} =-2 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+10}{2*4}=\frac{4}{8} =1/2 $
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