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

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

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

We multiply parentheses ..

(+3x^2-1x+15x-5)+(x-5)(2x+1)=0

We get rid of parentheses

3x^2-1x+15x+(x-5)(2x+1)-5=0

We multiply parentheses ..

3x^2+(+2x^2+x-10x-5)-1x+15x-5=0

We add all the numbers together, and all the variables

3x^2+(+2x^2+x-10x-5)+14x-5=0

We get rid of parentheses

3x^2+2x^2+x-10x+14x-5-5=0

We add all the numbers together, and all the variables

5x^2+5x-10=0

a = 5; b = 5; c = -10;
Δ = b2-4ac
Δ = 52-4·5·(-10)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-15}{2*5}=\frac{-20}{10}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+15}{2*5}=\frac{10}{10}
=1
$