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4a^2+20a+9=0
a = 4; b = 20; c = +9;
Δ = b2-4ac
Δ = 202-4·4·9
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{256}=16$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16}{2*4}=\frac{-36}{8} =-4+1/2 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16}{2*4}=\frac{-4}{8} =-1/2 $
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