(2x+5)(3x-2)=25

Solution for (2x+5)(3x-2)=25 equation:

(2x+5)(3x-2)=25

We move all terms to the left:

(2x+5)(3x-2)-(25)=0

We multiply parentheses ..

(+6x^2-4x+15x-10)-25=0

We get rid of parentheses

6x^2-4x+15x-10-25=0

We add all the numbers together, and all the variables

6x^2+11x-35=0

a = 6; b = 11; c = -35;
Δ = b2-4ac
Δ = 112-4·6·(-35)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-31}{2*6}=\frac{-42}{12}
=-3+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+31}{2*6}=\frac{20}{12}
=1+2/3
$