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5(x+2)2x+2=0
We multiply parentheses
10x^2+20x+2=0
a = 10; b = 20; c = +2;
Δ = b2-4ac
Δ = 202-4·10·2
Δ = 320
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8\sqrt{5}}{2*10}=\frac{-20-8\sqrt{5}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8\sqrt{5}}{2*10}=\frac{-20+8\sqrt{5}}{20} $
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