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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+6x^2-15x+4x-10)-7=0

We get rid of parentheses

6x^2-15x+4x-10-7=0

We add all the numbers together, and all the variables

6x^2-11x-17=0

a = 6; b = -11; c = -17;
Δ = b2-4ac
Δ = -112-4·6·(-17)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-23}{2*6}=\frac{-12}{12}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+23}{2*6}=\frac{34}{12}
=2+5/6
$