S^3+3s^2+7s+5=S^2+4s+3

Solution for S^3+3s^2+7s+5=S^2+4s+3 equation:

^3+3S^2+7S+5=^2+4S+3

We move all terms to the left:

^3+3S^2+7S+5-(^2+4S+3)=0

We add all the numbers together, and all the variables

3S^2+7S-(^2+4S+3)=0

We get rid of parentheses

3S^2+7S-4S-3-^2=0

We add all the numbers together, and all the variables

3S^2+3S=0

a = 3; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·3·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*3}=\frac{-6}{6}
=-1
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*3}=\frac{0}{6}
=0
$