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6m^2+25m+16=0
a = 6; b = 25; c = +16;
Δ = b2-4ac
Δ = 252-4·6·16
Δ = 241
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{-(25)-\sqrt{241}}{2*6}=\frac{-25-\sqrt{241}}{12} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{241}}{2*6}=\frac{-25+\sqrt{241}}{12} $
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