(7x+1)(5x+3)=136

Solution for (7x+1)(5x+3)=136 equation:

(7x+1)(5x+3)=136

We move all terms to the left:

(7x+1)(5x+3)-(136)=0

We multiply parentheses ..

(+35x^2+21x+5x+3)-136=0

We get rid of parentheses

35x^2+21x+5x+3-136=0

We add all the numbers together, and all the variables

35x^2+26x-133=0

a = 35; b = 26; c = -133;
Δ = b2-4ac
Δ = 262-4·35·(-133)
Δ = 19296
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:

$\sqrt{\Delta}=\sqrt{19296}=\sqrt{144*134}=\sqrt{144}*\sqrt{134}=12\sqrt{134}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-12\sqrt{134}}{2*35}=\frac{-26-12\sqrt{134}}{70}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+12\sqrt{134}}{2*35}=\frac{-26+12\sqrt{134}}{70}
$