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4m^2+m-95=0
a = 4; b = 1; c = -95;
Δ = b2-4ac
Δ = 12-4·4·(-95)
Δ = 1521
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{1521}=39$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-39}{2*4}=\frac{-40}{8} =-5 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+39}{2*4}=\frac{38}{8} =4+3/4 $
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