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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We multiply parentheses ..

2x^2+(+2x^2-4x+x-2)+10x-1x-5=0

We add all the numbers together, and all the variables

2x^2+(+2x^2-4x+x-2)+9x-5=0

We get rid of parentheses

2x^2+2x^2-4x+x+9x-2-5=0

We add all the numbers together, and all the variables

4x^2+6x-7=0

a = 4; b = 6; c = -7;
Δ = b2-4ac
Δ = 62-4·4·(-7)
Δ = 148
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{37}}{2*4}=\frac{-6-2\sqrt{37}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{37}}{2*4}=\frac{-6+2\sqrt{37}}{8}
$