2(5x+4)+(x^2+x-2)=0

Solution for 2(5x+4)+(x^2+x-2)=0 equation:

2(5x+4)+(x^2+x-2)=0

We multiply parentheses

10x+(x^2+x-2)+8=0

We get rid of parentheses

x^2+10x+x-2+8=0

We add all the numbers together, and all the variables

x^2+11x+6=0

a = 1; b = 11; c = +6;
Δ = b2-4ac
Δ = 112-4·1·6
Δ = 97
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{-(11)-\sqrt{97}}{2*1}=\frac{-11-\sqrt{97}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{97}}{2*1}=\frac{-11+\sqrt{97}}{2}
$