x^2-4+x=0

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Solution for x^2-4+x=0 equation: 



x^2-4+x=0

a = 1; b = 1; c = -4;
Δ = b2-4ac
Δ = 12-4·1·(-4)
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{17}}{2*1}=\frac{-1-\sqrt{17}}{2}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{17}}{2*1}=\frac{-1+\sqrt{17}}{2}
$

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