(5x+2)(3x+4)=(7x+7)

Solution for (5x+2)(3x+4)=(7x+7) equation:

(5x+2)(3x+4)=(7x+7)

We move all terms to the left:

(5x+2)(3x+4)-((7x+7))=0

We multiply parentheses ..

(+15x^2+20x+6x+8)-((7x+7))=0


We calculate terms in parentheses: -((7x+7)), so:

(7x+7)

We get rid of parentheses

7x+7

Back to the equation:

-(7x+7)


We get rid of parentheses

15x^2+20x+6x-7x+8-7=0

We add all the numbers together, and all the variables

15x^2+19x+1=0

a = 15; b = 19; c = +1;
Δ = b2-4ac
Δ = 192-4·15·1
Δ = 301
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{-(19)-\sqrt{301}}{2*15}=\frac{-19-\sqrt{301}}{30}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{301}}{2*15}=\frac{-19+\sqrt{301}}{30}
$