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

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

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

We move all terms to the left:

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


We calculate terms in parentheses: -((3x+5)2x), so:

(3x+5)2x

We multiply parentheses

6x^2+10x

Back to the equation:

-(6x^2+10x)


We get rid of parentheses

-6x^2-10x+40=0

a = -6; b = -10; c = +40;
Δ = b2-4ac
Δ = -102-4·(-6)·40
Δ = 1060
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:

$\sqrt{\Delta}=\sqrt{1060}=\sqrt{4*265}=\sqrt{4}*\sqrt{265}=2\sqrt{265}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{265}}{2*-6}=\frac{10-2\sqrt{265}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{265}}{2*-6}=\frac{10+2\sqrt{265}}{-12}
$