180=(3x+5)(10x-7)

Solution for 180=(3x+5)(10x-7) equation:

180=(3x+5)(10x-7)

We move all terms to the left:

180-((3x+5)(10x-7))=0

We multiply parentheses ..

-((+30x^2-21x+50x-35))+180=0


We calculate terms in parentheses: -((+30x^2-21x+50x-35)), so:

(+30x^2-21x+50x-35)

We get rid of parentheses

30x^2-21x+50x-35

We add all the numbers together, and all the variables

30x^2+29x-35

Back to the equation:

-(30x^2+29x-35)


We get rid of parentheses

-30x^2-29x+35+180=0

We add all the numbers together, and all the variables

-30x^2-29x+215=0

a = -30; b = -29; c = +215;
Δ = b2-4ac
Δ = -292-4·(-30)·215
Δ = 26641
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-\sqrt{26641}}{2*-30}=\frac{29-\sqrt{26641}}{-60}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+\sqrt{26641}}{2*-30}=\frac{29+\sqrt{26641}}{-60}
$