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4m^2+17m-15=0
a = 4; b = 17; c = -15;
Δ = b2-4ac
Δ = 172-4·4·(-15)
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$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-23}{2*4}=\frac{-40}{8} =-5 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+23}{2*4}=\frac{6}{8} =3/4 $
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