1=(x^2+2x-5)

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Solution for 1=(x^2+2x-5) equation:



1=(x^2+2x-5)
We move all terms to the left:
1-((x^2+2x-5))=0
We calculate terms in parentheses: -((x^2+2x-5)), so:
(x^2+2x-5)
We get rid of parentheses
x^2+2x-5
Back to the equation:
-(x^2+2x-5)
We get rid of parentheses
-x^2-2x+5+1=0
We add all the numbers together, and all the variables
-1x^2-2x+6=0
a = -1; b = -2; c = +6;
Δ = b2-4ac
Δ = -22-4·(-1)·6
Δ = 28
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:
$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{7}}{2*-1}=\frac{2-2\sqrt{7}}{-2} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{7}}{2*-1}=\frac{2+2\sqrt{7}}{-2} $

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