3n2+10n=25

Solution for 3n2+10n=25 equation:

3n^2+10n=25

We move all terms to the left:

3n^2+10n-(25)=0

a = 3; b = 10; c = -25;
Δ = b2-4ac
Δ = 102-4·3·(-25)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-20}{2*3}=\frac{-30}{6}
=-5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+20}{2*3}=\frac{10}{6}
=1+2/3
$