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