5n^2+19n-42=9n+9

Solution for 5n^2+19n-42=9n+9 equation:

5n^2+19n-42=9n+9

We move all terms to the left:

5n^2+19n-42-(9n+9)=0

We get rid of parentheses

5n^2+19n-9n-9-42=0

We add all the numbers together, and all the variables

5n^2+10n-51=0

a = 5; b = 10; c = -51;
Δ = b2-4ac
Δ = 102-4·5·(-51)
Δ = 1120
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:

$\sqrt{\Delta}=\sqrt{1120}=\sqrt{16*70}=\sqrt{16}*\sqrt{70}=4\sqrt{70}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{70}}{2*5}=\frac{-10-4\sqrt{70}}{10}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{70}}{2*5}=\frac{-10+4\sqrt{70}}{10}
$