(2n2-5n-6)+(-n2-3n+11)=0

Solution for (2n2-5n-6)+(-n2-3n+11)=0 equation:

(2n^2-5n-6)+(-n2-3n+11)=0

We add all the numbers together, and all the variables

(-1n^2-3n+11)+(2n^2-5n-6)=0

We get rid of parentheses

-1n^2+2n^2-3n-5n+11-6=0

We add all the numbers together, and all the variables

n^2-8n+5=0

a = 1; b = -8; c = +5;
Δ = b2-4ac
Δ = -82-4·1·5
Δ = 44
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{11}}{2*1}=\frac{8-2\sqrt{11}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{11}}{2*1}=\frac{8+2\sqrt{11}}{2}
$