(2x+5)(3x-2)=(x+1)

Solution for (2x+5)(3x-2)=(x+1) equation:

(2x+5)(3x-2)=(x+1)

We move all terms to the left:

(2x+5)(3x-2)-((x+1))=0

We multiply parentheses ..

(+6x^2-4x+15x-10)-((x+1))=0


We calculate terms in parentheses: -((x+1)), so:

(x+1)

We get rid of parentheses

x+1

Back to the equation:

-(x+1)


We get rid of parentheses

6x^2-4x+15x-x-10-1=0

We add all the numbers together, and all the variables

6x^2+10x-11=0

a = 6; b = 10; c = -11;
Δ = b2-4ac
Δ = 102-4·6·(-11)
Δ = 364
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:

$\sqrt{\Delta}=\sqrt{364}=\sqrt{4*91}=\sqrt{4}*\sqrt{91}=2\sqrt{91}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{91}}{2*6}=\frac{-10-2\sqrt{91}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{91}}{2*6}=\frac{-10+2\sqrt{91}}{12}
$