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5x^2+20x+2=0
a = 5; b = 20; c = +2;
Δ = b2-4ac
Δ = 202-4·5·2
Δ = 360
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-6\sqrt{10}}{2*5}=\frac{-20-6\sqrt{10}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+6\sqrt{10}}{2*5}=\frac{-20+6\sqrt{10}}{10} $
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