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5m^2-26m+33=0
a = 5; b = -26; c = +33;
Δ = b2-4ac
Δ = -262-4·5·33
Δ = 16
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{16}=4$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-4}{2*5}=\frac{22}{10} =2+1/5 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+4}{2*5}=\frac{30}{10} =3 $
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