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2x+x^2=450
We move all terms to the left:
2x+x^2-(450)=0
a = 1; b = 2; c = -450;
Δ = b2-4ac
Δ = 22-4·1·(-450)
Δ = 1804
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1804}=\sqrt{4*451}=\sqrt{4}*\sqrt{451}=2\sqrt{451}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{451}}{2*1}=\frac{-2-2\sqrt{451}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{451}}{2*1}=\frac{-2+2\sqrt{451}}{2} $
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