5x(7x+5)+(7x+4)=161

Solution for 5x(7x+5)+(7x+4)=161 equation:

5x(7x+5)+(7x+4)=161

We move all terms to the left:

5x(7x+5)+(7x+4)-(161)=0

We multiply parentheses

35x^2+25x+(7x+4)-161=0

We get rid of parentheses

35x^2+25x+7x+4-161=0

We add all the numbers together, and all the variables

35x^2+32x-157=0

a = 35; b = 32; c = -157;
Δ = b2-4ac
Δ = 322-4·35·(-157)
Δ = 23004
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:

$\sqrt{\Delta}=\sqrt{23004}=\sqrt{324*71}=\sqrt{324}*\sqrt{71}=18\sqrt{71}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-18\sqrt{71}}{2*35}=\frac{-32-18\sqrt{71}}{70}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+18\sqrt{71}}{2*35}=\frac{-32+18\sqrt{71}}{70}
$