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

Solution for (x-3)*(2x+5)-4*(x-3)=0 equation:

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

We multiply parentheses

(x-3)(2x+5)-4x+12=0

We multiply parentheses ..

(+2x^2+5x-6x-15)-4x+12=0

We get rid of parentheses

2x^2+5x-6x-4x-15+12=0

We add all the numbers together, and all the variables

2x^2-5x-3=0

a = 2; b = -5; c = -3;
Δ = b2-4ac
Δ = -52-4·2·(-3)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-7}{2*2}=\frac{-2}{4}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+7}{2*2}=\frac{12}{4}
=3
$