4n2+11n=15

Solution for 4n2+11n=15 equation:

4n^2+11n=15

We move all terms to the left:

4n^2+11n-(15)=0

a = 4; b = 11; c = -15;
Δ = b2-4ac
Δ = 112-4·4·(-15)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-19}{2*4}=\frac{-30}{8}
=-3+3/4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+19}{2*4}=\frac{8}{8}
=1
$