(3e+10)+(5+5e)e=10

Solution for (3e+10)+(5+5e)e=10 equation:

(3e+10)+(5+5e)e=10

We move all terms to the left:

(3e+10)+(5+5e)e-(10)=0

We add all the numbers together, and all the variables

(3e+10)+(5e+5)e-10=0

We multiply parentheses

5e^2+(3e+10)+5e-10=0

We get rid of parentheses

5e^2+3e+5e+10-10=0

We add all the numbers together, and all the variables

5e^2+8e=0

a = 5; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·5·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*5}=\frac{-16}{10}
=-1+3/5
$
$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*5}=\frac{0}{10}
=0
$