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2x-x^2+16=0
We add all the numbers together, and all the variables
-1x^2+2x+16=0
a = -1; b = 2; c = +16;
Δ = b2-4ac
Δ = 22-4·(-1)·16
Δ = 68
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{17}}{2*-1}=\frac{-2-2\sqrt{17}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{17}}{2*-1}=\frac{-2+2\sqrt{17}}{-2} $
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