5n^2-22n-48=0

Solution for 5n^2-22n-48=0 equation:

5n^2-22n-48=0

a = 5; b = -22; c = -48;
Δ = b2-4ac
Δ = -222-4·5·(-48)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-38}{2*5}=\frac{-16}{10}
=-1+3/5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+38}{2*5}=\frac{60}{10}
=6
$