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4m^2+25m+6=0
a = 4; b = 25; c = +6;
Δ = b2-4ac
Δ = 252-4·4·6
Δ = 529
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{529}=23$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-23}{2*4}=\frac{-48}{8} =-6 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+23}{2*4}=\frac{-2}{8} =-1/4 $
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