2n^2-n-4=2+5

Solution for 2n^2-n-4=2+5 equation:

2n^2-n-4=2+5

We move all terms to the left:

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

We add all the numbers together, and all the variables

2n^2-n-4-7=0

We add all the numbers together, and all the variables

2n^2-1n-11=0

a = 2; b = -1; c = -11;
Δ = b2-4ac
Δ = -12-4·2·(-11)
Δ = 89
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{89}}{2*2}=\frac{1-\sqrt{89}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{89}}{2*2}=\frac{1+\sqrt{89}}{4}
$