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4m^2-3m=1
We move all terms to the left:
4m^2-3m-(1)=0
a = 4; b = -3; c = -1;
Δ = b2-4ac
Δ = -32-4·4·(-1)
Δ = 25
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{25}=5$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2*4}=\frac{-2}{8} =-1/4 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*4}=\frac{8}{8} =1 $
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