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10w^2-13w+4=0
a = 10; b = -13; c = +4;
Δ = b2-4ac
Δ = -132-4·10·4
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{9}=3$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-3}{2*10}=\frac{10}{20} =1/2 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+3}{2*10}=\frac{16}{20} =4/5 $
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