(x-5)(x+10)=(x+1)2

Solution for (x-5)(x+10)=(x+1)2 equation:

(x-5)(x+10)=(x+1)2

We move all terms to the left:

(x-5)(x+10)-((x+1)2)=0

We multiply parentheses ..

(+x^2+10x-5x-50)-((x+1)2)=0


We calculate terms in parentheses: -((x+1)2), so:

(x+1)2

We multiply parentheses

2x+2

Back to the equation:

-(2x+2)


We get rid of parentheses

x^2+10x-5x-2x-50-2=0

We add all the numbers together, and all the variables

x^2+3x-52=0

a = 1; b = 3; c = -52;
Δ = b2-4ac
Δ = 32-4·1·(-52)
Δ = 217
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{-(3)-\sqrt{217}}{2*1}=\frac{-3-\sqrt{217}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{217}}{2*1}=\frac{-3+\sqrt{217}}{2}
$