(x+40)=(3x-17)(2x-5)

Solution for (x+40)=(3x-17)(2x-5) equation:

(x+40)=(3x-17)(2x-5)

We move all terms to the left:

(x+40)-((3x-17)(2x-5))=0

We get rid of parentheses

x-((3x-17)(2x-5))+40=0

We multiply parentheses ..

-((+6x^2-15x-34x+85))+x+40=0


We calculate terms in parentheses: -((+6x^2-15x-34x+85)), so:

(+6x^2-15x-34x+85)

We get rid of parentheses

6x^2-15x-34x+85

We add all the numbers together, and all the variables

6x^2-49x+85

Back to the equation:

-(6x^2-49x+85)


We add all the numbers together, and all the variables

x-(6x^2-49x+85)+40=0

We get rid of parentheses

-6x^2+x+49x-85+40=0

We add all the numbers together, and all the variables

-6x^2+50x-45=0

a = -6; b = 50; c = -45;
Δ = b2-4ac
Δ = 502-4·(-6)·(-45)
Δ = 1420
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:

$\sqrt{\Delta}=\sqrt{1420}=\sqrt{4*355}=\sqrt{4}*\sqrt{355}=2\sqrt{355}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{355}}{2*-6}=\frac{-50-2\sqrt{355}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{355}}{2*-6}=\frac{-50+2\sqrt{355}}{-12}
$