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20q^2-90q+40=0
a = 20; b = -90; c = +40;
Δ = b2-4ac
Δ = -902-4·20·40
Δ = 4900
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{4900}=70$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-90)-70}{2*20}=\frac{20}{40} =1/2 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-90)+70}{2*20}=\frac{160}{40} =4 $
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