(4x-1)(x+3)+(4x-1)(x+6)=0

Solution for (4x-1)(x+3)+(4x-1)(x+6)=0 equation:

(4x-1)(x+3)+(4x-1)(x+6)=0

We multiply parentheses ..

(+4x^2+12x-1x-3)+(4x-1)(x+6)=0

We get rid of parentheses

4x^2+12x-1x+(4x-1)(x+6)-3=0

We multiply parentheses ..

4x^2+(+4x^2+24x-1x-6)+12x-1x-3=0

We add all the numbers together, and all the variables

4x^2+(+4x^2+24x-1x-6)+11x-3=0

We get rid of parentheses

4x^2+4x^2+24x-1x+11x-6-3=0

We add all the numbers together, and all the variables

8x^2+34x-9=0

a = 8; b = 34; c = -9;
Δ = b2-4ac
Δ = 342-4·8·(-9)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-38}{2*8}=\frac{-72}{16}
=-4+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+38}{2*8}=\frac{4}{16}
=1/4
$