10x^2+x(1+x)=280

Solution for 10x^2+x(1+x)=280 equation:

10x^2+x(1+x)=280

We move all terms to the left:

10x^2+x(1+x)-(280)=0

We add all the numbers together, and all the variables

10x^2+x(x+1)-280=0

We multiply parentheses

10x^2+x^2+x-280=0

We add all the numbers together, and all the variables

11x^2+x-280=0

a = 11; b = 1; c = -280;
Δ = b2-4ac
Δ = 12-4·11·(-280)
Δ = 12321
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12321}=111$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-111}{2*11}=\frac{-112}{22}
=-5+1/11
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+111}{2*11}=\frac{110}{22}
=5
$