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m^2+15m+4=0
a = 1; b = 15; c = +4;
Δ = b2-4ac
Δ = 152-4·1·4
Δ = 209
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{-(15)-\sqrt{209}}{2*1}=\frac{-15-\sqrt{209}}{2} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{209}}{2*1}=\frac{-15+\sqrt{209}}{2} $
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