2s+4=5(-4-2s)2s+4

Solution for 2s+4=5(-4-2s)2s+4 equation:

2s+4=5(-4-2s)2s+4

We move all terms to the left:

2s+4-(5(-4-2s)2s+4)=0

We add all the numbers together, and all the variables

2s-(5(-2s-4)2s+4)+4=0


We calculate terms in parentheses: -(5(-2s-4)2s+4), so:

5(-2s-4)2s+4

We multiply parentheses

-20s^2-40s+4

Back to the equation:

-(-20s^2-40s+4)


We get rid of parentheses

20s^2+40s+2s-4+4=0

We add all the numbers together, and all the variables

20s^2+42s=0

a = 20; b = 42; c = 0;
Δ = b2-4ac
Δ = 422-4·20·0
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-42}{2*20}=\frac{-84}{40}
=-2+1/10
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+42}{2*20}=\frac{0}{40}
=0
$