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4m^2-3m=2
We move all terms to the left:
4m^2-3m-(2)=0
a = 4; b = -3; c = -2;
Δ = b2-4ac
Δ = -32-4·4·(-2)
Δ = 41
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{41}}{2*4}=\frac{3-\sqrt{41}}{8} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{41}}{2*4}=\frac{3+\sqrt{41}}{8} $
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