(2x+1)(10x+2)=155

Solution for (2x+1)(10x+2)=155 equation:

(2x+1)(10x+2)=155

We move all terms to the left:

(2x+1)(10x+2)-(155)=0

We multiply parentheses ..

(+20x^2+4x+10x+2)-155=0

We get rid of parentheses

20x^2+4x+10x+2-155=0

We add all the numbers together, and all the variables

20x^2+14x-153=0

a = 20; b = 14; c = -153;
Δ = b2-4ac
Δ = 142-4·20·(-153)
Δ = 12436
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:

$\sqrt{\Delta}=\sqrt{12436}=\sqrt{4*3109}=\sqrt{4}*\sqrt{3109}=2\sqrt{3109}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{3109}}{2*20}=\frac{-14-2\sqrt{3109}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{3109}}{2*20}=\frac{-14+2\sqrt{3109}}{40}
$