s+(2+1/2)s+20=62

Solution for s+(2+1/2)s+20=62 equation:

s+(2+1/2)s+20=62

We move all terms to the left:

s+(2+1/2)s+20-(62)=0


Domain of the equation: 2)s!=0

s!=0/1

s!=0

s∈R


We add all the numbers together, and all the variables

s+(1/2+2)s+20-62=0

We add all the numbers together, and all the variables

s+(1/2+2)s-42=0

We multiply parentheses

s^2+s+2s-42=0

We add all the numbers together, and all the variables

s^2+3s-42=0

a = 1; b = 3; c = -42;
Δ = b2-4ac
Δ = 32-4·1·(-42)
Δ = 177
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{177}}{2*1}=\frac{-3-\sqrt{177}}{2}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{177}}{2*1}=\frac{-3+\sqrt{177}}{2}
$