40+5x=5(7+x2x)

Solution for 40+5x=5(7+x2x) equation:

40+5x=5(7+x2x)

We move all terms to the left:

40+5x-(5(7+x2x))=0

We add all the numbers together, and all the variables

5x-(5(x2x+7))+40=0


We calculate terms in parentheses: -(5(x2x+7)), so:

5(x2x+7)

We multiply parentheses

5x^2+35

Back to the equation:

-(5x^2+35)


We get rid of parentheses

-5x^2+5x-35+40=0

We add all the numbers together, and all the variables

-5x^2+5x+5=0

a = -5; b = 5; c = +5;
Δ = b2-4ac
Δ = 52-4·(-5)·5
Δ = 125
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{125}=\sqrt{25*5}=\sqrt{25}*\sqrt{5}=5\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{5}}{2*-5}=\frac{-5-5\sqrt{5}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{5}}{2*-5}=\frac{-5+5\sqrt{5}}{-10}
$