(3x-5)(7x+9)=21x+4

Solution for (3x-5)(7x+9)=21x+4 equation:

(3x-5)(7x+9)=21x+4

We move all terms to the left:

(3x-5)(7x+9)-(21x+4)=0

We get rid of parentheses

(3x-5)(7x+9)-21x-4=0

We multiply parentheses ..

(+21x^2+27x-35x-45)-21x-4=0

We get rid of parentheses

21x^2+27x-35x-21x-45-4=0

We add all the numbers together, and all the variables

21x^2-29x-49=0

a = 21; b = -29; c = -49;
Δ = b2-4ac
Δ = -292-4·21·(-49)
Δ = 4957
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-\sqrt{4957}}{2*21}=\frac{29-\sqrt{4957}}{42}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+\sqrt{4957}}{2*21}=\frac{29+\sqrt{4957}}{42}
$