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(x^2+10x)+5=0
We get rid of parentheses
x^2+10x+5=0
a = 1; b = 10; c = +5;
Δ = b2-4ac
Δ = 102-4·1·5
Δ = 80
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{5}}{2*1}=\frac{-10-4\sqrt{5}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{5}}{2*1}=\frac{-10+4\sqrt{5}}{2} $
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