(x+3)(2x-5)=(x+3)(x-4)

Solution for (x+3)(2x-5)=(x+3)(x-4) equation:

(x+3)(2x-5)=(x+3)(x-4)

We move all terms to the left:

(x+3)(2x-5)-((x+3)(x-4))=0

We multiply parentheses ..

(+2x^2-5x+6x-15)-((x+3)(x-4))=0


We calculate terms in parentheses: -((x+3)(x-4)), so:

(x+3)(x-4)

We multiply parentheses ..

(+x^2-4x+3x-12)

We get rid of parentheses

x^2-4x+3x-12

We add all the numbers together, and all the variables

x^2-1x-12

Back to the equation:

-(x^2-1x-12)


We get rid of parentheses

2x^2-x^2-5x+6x+1x-15+12=0

We add all the numbers together, and all the variables

x^2+2x-3=0

a = 1; b = 2; c = -3;
Δ = b2-4ac
Δ = 22-4·1·(-3)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4}{2*1}=\frac{-6}{2}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4}{2*1}=\frac{2}{2}
=1
$