5(x+5)=40x+5(x^2+11)

Solution for 5(x+5)=40x+5(x^2+11) equation:

5(x+5)=40x+5(x^2+11)

We move all terms to the left:

5(x+5)-(40x+5(x^2+11))=0

We multiply parentheses

5x-(40x+5(x^2+11))+25=0


We calculate terms in parentheses: -(40x+5(x^2+11)), so:

40x+5(x^2+11)

We multiply parentheses

5x^2+40x+55

Back to the equation:

-(5x^2+40x+55)


We get rid of parentheses

-5x^2+5x-40x-55+25=0

We add all the numbers together, and all the variables

-5x^2-35x-30=0

a = -5; b = -35; c = -30;
Δ = b2-4ac
Δ = -352-4·(-5)·(-30)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-25}{2*-5}=\frac{10}{-10}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+25}{2*-5}=\frac{60}{-10}
=-6
$