(x-3)(x+2)+(2x-1)(x+5)=0

Solution for (x-3)(x+2)+(2x-1)(x+5)=0 equation:

(x-3)(x+2)+(2x-1)(x+5)=0

We multiply parentheses ..

(+x^2+2x-3x-6)+(2x-1)(x+5)=0

We get rid of parentheses

x^2+2x-3x+(2x-1)(x+5)-6=0

We multiply parentheses ..

x^2+(+2x^2+10x-1x-5)+2x-3x-6=0

We add all the numbers together, and all the variables

x^2+(+2x^2+10x-1x-5)-1x-6=0

We get rid of parentheses

x^2+2x^2+10x-1x-1x-5-6=0

We add all the numbers together, and all the variables

3x^2+8x-11=0

a = 3; b = 8; c = -11;
Δ = b2-4ac
Δ = 82-4·3·(-11)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-14}{2*3}=\frac{-22}{6}
=-3+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+14}{2*3}=\frac{6}{6}
=1
$