(4x+3)=(3x+5)(4x-1)

Solution for (4x+3)=(3x+5)(4x-1) equation:

(4x+3)=(3x+5)(4x-1)

We move all terms to the left:

(4x+3)-((3x+5)(4x-1))=0

We get rid of parentheses

4x-((3x+5)(4x-1))+3=0

We multiply parentheses ..

-((+12x^2-3x+20x-5))+4x+3=0


We calculate terms in parentheses: -((+12x^2-3x+20x-5)), so:

(+12x^2-3x+20x-5)

We get rid of parentheses

12x^2-3x+20x-5

We add all the numbers together, and all the variables

12x^2+17x-5

Back to the equation:

-(12x^2+17x-5)


We add all the numbers together, and all the variables

4x-(12x^2+17x-5)+3=0

We get rid of parentheses

-12x^2+4x-17x+5+3=0

We add all the numbers together, and all the variables

-12x^2-13x+8=0

a = -12; b = -13; c = +8;
Δ = b2-4ac
Δ = -132-4·(-12)·8
Δ = 553
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{-(-13)-\sqrt{553}}{2*-12}=\frac{13-\sqrt{553}}{-24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{553}}{2*-12}=\frac{13+\sqrt{553}}{-24}
$