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2x+x2=170
We move all terms to the left:
2x+x2-(170)=0
We add all the numbers together, and all the variables
x^2+2x-170=0
a = 1; b = 2; c = -170;
Δ = b2-4ac
Δ = 22-4·1·(-170)
Δ = 684
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{684}=\sqrt{36*19}=\sqrt{36}*\sqrt{19}=6\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6\sqrt{19}}{2*1}=\frac{-2-6\sqrt{19}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6\sqrt{19}}{2*1}=\frac{-2+6\sqrt{19}}{2} $
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