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m^2+14m-15=0
a = 1; b = 14; c = -15;
Δ = b2-4ac
Δ = 142-4·1·(-15)
Δ = 256
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{256}=16$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-16}{2*1}=\frac{-30}{2} =-15 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+16}{2*1}=\frac{2}{2} =1 $
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