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

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}
$