(3x-10)(4x=40)

Solution for (3x-10)(4x=40) equation:

(3x-10)(4x=40)

We move all terms to the left:

(3x-10)(4x-(40))=0

We multiply parentheses ..

(+12x^2-120x-40x+400)=0

We get rid of parentheses

12x^2-120x-40x+400=0

We add all the numbers together, and all the variables

12x^2-160x+400=0

a = 12; b = -160; c = +400;
Δ = b2-4ac
Δ = -1602-4·12·400
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-80}{2*12}=\frac{80}{24}
=3+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+80}{2*12}=\frac{240}{24}
=10
$