7*(5e+2e*4e)=3+e

Solution for 7*(5e+2e*4e)=3+e equation:

7(5e+2e*4e)=3+e

We move all terms to the left:

7(5e+2e*4e)-(3+e)=0

We add all the numbers together, and all the variables

7(+5e+2e*4e)-(e+3)=0

We multiply parentheses

56e^2+35e-(e+3)=0

We get rid of parentheses

56e^2+35e-e-3=0

We add all the numbers together, and all the variables

56e^2+34e-3=0

a = 56; b = 34; c = -3;
Δ = b2-4ac
Δ = 342-4·56·(-3)
Δ = 1828
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1828}=\sqrt{4*457}=\sqrt{4}*\sqrt{457}=2\sqrt{457}$
$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-2\sqrt{457}}{2*56}=\frac{-34-2\sqrt{457}}{112}
$
$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+2\sqrt{457}}{2*56}=\frac{-34+2\sqrt{457}}{112}
$