2n^2-26n+48=0

Solution for 2n^2-26n+48=0 equation:

2n^2-26n+48=0

a = 2; b = -26; c = +48;
Δ = b2-4ac
Δ = -262-4·2·48
Δ = 292
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:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2\sqrt{73}}{2*2}=\frac{26-2\sqrt{73}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2\sqrt{73}}{2*2}=\frac{26+2\sqrt{73}}{4}
$