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x^2+2x=293
We move all terms to the left:
x^2+2x-(293)=0
a = 1; b = 2; c = -293;
Δ = b2-4ac
Δ = 22-4·1·(-293)
Δ = 1176
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1176}=\sqrt{196*6}=\sqrt{196}*\sqrt{6}=14\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14\sqrt{6}}{2*1}=\frac{-2-14\sqrt{6}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14\sqrt{6}}{2*1}=\frac{-2+14\sqrt{6}}{2} $
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