5n^2+29n-42=0

Solution for 5n^2+29n-42=0 equation:

5n^2+29n-42=0

a = 5; b = 29; c = -42;
Δ = b2-4ac
Δ = 292-4·5·(-42)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-41}{2*5}=\frac{-70}{10}
=-7
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+41}{2*5}=\frac{12}{10}
=1+1/5
$