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4x^2+20x+4=0
a = 4; b = 20; c = +4;
Δ = b2-4ac
Δ = 202-4·4·4
Δ = 336
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{21}}{2*4}=\frac{-20-4\sqrt{21}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{21}}{2*4}=\frac{-20+4\sqrt{21}}{8} $
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