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5x+20x^2=0
a = 20; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·20·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*20}=\frac{-10}{40} =-1/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*20}=\frac{0}{40} =0 $
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