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9x^2+20x+4=0
a = 9; b = 20; c = +4;
Δ = b2-4ac
Δ = 202-4·9·4
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16}{2*9}=\frac{-36}{18} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16}{2*9}=\frac{-4}{18} =-2/9 $
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