(x+40)(4x-5)=(6x+20)

Solution for (x+40)(4x-5)=(6x+20) equation:

(x+40)(4x-5)=(6x+20)

We move all terms to the left:

(x+40)(4x-5)-((6x+20))=0

We multiply parentheses ..

(+4x^2-5x+160x-200)-((6x+20))=0


We calculate terms in parentheses: -((6x+20)), so:

(6x+20)

We get rid of parentheses

6x+20

Back to the equation:

-(6x+20)


We get rid of parentheses

4x^2-5x+160x-6x-200-20=0

We add all the numbers together, and all the variables

4x^2+149x-220=0

a = 4; b = 149; c = -220;
Δ = b2-4ac
Δ = 1492-4·4·(-220)
Δ = 25721
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:

$\sqrt{\Delta}=\sqrt{25721}=\sqrt{289*89}=\sqrt{289}*\sqrt{89}=17\sqrt{89}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(149)-17\sqrt{89}}{2*4}=\frac{-149-17\sqrt{89}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(149)+17\sqrt{89}}{2*4}=\frac{-149+17\sqrt{89}}{8}
$