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.33m^2+4m+9=0
a = .33; b = 4; c = +9;
Δ = b2-4ac
Δ = 42-4·.33·9
Δ = 4.12
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{-(4)-\sqrt{4.12}}{2*.33}=\frac{-4-\sqrt{4.12}}{0.66} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+\sqrt{4.12}}{2*.33}=\frac{-4+\sqrt{4.12}}{0.66} $
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