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m^2+12m+17=0
a = 1; b = 12; c = +17;
Δ = b2-4ac
Δ = 122-4·1·17
Δ = 76
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{19}}{2*1}=\frac{-12-2\sqrt{19}}{2} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{19}}{2*1}=\frac{-12+2\sqrt{19}}{2} $
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