2x(3x+5)=41

Solution for 2x(3x+5)=41 equation:

2x(3x+5)=41

We move all terms to the left:

2x(3x+5)-(41)=0

We multiply parentheses

6x^2+10x-41=0

a = 6; b = 10; c = -41;
Δ = b2-4ac
Δ = 102-4·6·(-41)
Δ = 1084
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:

$\sqrt{\Delta}=\sqrt{1084}=\sqrt{4*271}=\sqrt{4}*\sqrt{271}=2\sqrt{271}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{271}}{2*6}=\frac{-10-2\sqrt{271}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{271}}{2*6}=\frac{-10+2\sqrt{271}}{12}
$