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

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

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

We move all terms to the left:

(x+1)(2x-5)-(1-x^2)=0

We get rid of parentheses

x^2+(x+1)(2x-5)-1=0

We multiply parentheses ..

x^2+(+2x^2-5x+2x-5)-1=0

We get rid of parentheses

x^2+2x^2-5x+2x-5-1=0

We add all the numbers together, and all the variables

3x^2-3x-6=0

a = 3; b = -3; c = -6;
Δ = b2-4ac
Δ = -32-4·3·(-6)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-9}{2*3}=\frac{-6}{6}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+9}{2*3}=\frac{12}{6}
=2
$