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4x^2+20x+3=0
a = 4; b = 20; c = +3;
Δ = b2-4ac
Δ = 202-4·4·3
Δ = 352
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{352}=\sqrt{16*22}=\sqrt{16}*\sqrt{22}=4\sqrt{22}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{22}}{2*4}=\frac{-20-4\sqrt{22}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{22}}{2*4}=\frac{-20+4\sqrt{22}}{8} $
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