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5m^2-3m-15=0
a = 5; b = -3; c = -15;
Δ = b2-4ac
Δ = -32-4·5·(-15)
Δ = 309
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{-(-3)-\sqrt{309}}{2*5}=\frac{3-\sqrt{309}}{10} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{309}}{2*5}=\frac{3+\sqrt{309}}{10} $
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