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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+7x^2+5x+21x+15))+1=0


We calculate terms in parentheses: -((+7x^2+5x+21x+15)), so:

(+7x^2+5x+21x+15)

We get rid of parentheses

7x^2+5x+21x+15

We add all the numbers together, and all the variables

7x^2+26x+15

Back to the equation:

-(7x^2+26x+15)


We get rid of parentheses

-7x^2-26x-15+1=0

We add all the numbers together, and all the variables

-7x^2-26x-14=0

a = -7; b = -26; c = -14;
Δ = b2-4ac
Δ = -262-4·(-7)·(-14)
Δ = 284
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:

$\sqrt{\Delta}=\sqrt{284}=\sqrt{4*71}=\sqrt{4}*\sqrt{71}=2\sqrt{71}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2\sqrt{71}}{2*-7}=\frac{26-2\sqrt{71}}{-14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2\sqrt{71}}{2*-7}=\frac{26+2\sqrt{71}}{-14}
$