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

Solution for (x-3)(2x+5)+(x-3)(x-7)=0 equation:

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

We multiply parentheses ..

(+2x^2+5x-6x-15)+(x-3)(x-7)=0

We get rid of parentheses

2x^2+5x-6x+(x-3)(x-7)-15=0

We multiply parentheses ..

2x^2+(+x^2-7x-3x+21)+5x-6x-15=0

We add all the numbers together, and all the variables

2x^2+(+x^2-7x-3x+21)-1x-15=0

We get rid of parentheses

2x^2+x^2-7x-3x-1x+21-15=0

We add all the numbers together, and all the variables

3x^2-11x+6=0

a = 3; b = -11; c = +6;
Δ = b2-4ac
Δ = -112-4·3·6
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{-(-11)-7}{2*3}=\frac{4}{6}
=2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+7}{2*3}=\frac{18}{6}
=3
$