(4x-1)(2x+3)=(x+3)(x1)

Solution for (4x-1)(2x+3)=(x+3)(x1) equation:

(4x-1)(2x+3)=(x+3)(x1)

We move all terms to the left:

(4x-1)(2x+3)-((x+3)(x1))=0

We multiply parentheses ..

(+8x^2+12x-2x-3)-((x+3)x1)=0


We calculate terms in parentheses: -((x+3)x1), so:

(x+3)x1

We multiply parentheses

x^2+3x

Back to the equation:

-(x^2+3x)


We get rid of parentheses

8x^2-x^2+12x-2x-3x-3=0

We add all the numbers together, and all the variables

7x^2+7x-3=0

a = 7; b = 7; c = -3;
Δ = b2-4ac
Δ = 72-4·7·(-3)
Δ = 133
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{-(7)-\sqrt{133}}{2*7}=\frac{-7-\sqrt{133}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{133}}{2*7}=\frac{-7+\sqrt{133}}{14}
$