(2x+3)(9x-8)+(2x+3)(10x-2)=0

Solution for (2x+3)(9x-8)+(2x+3)(10x-2)=0 equation:

(2x+3)(9x-8)+(2x+3)(10x-2)=0

We multiply parentheses ..

(+18x^2-16x+27x-24)+(2x+3)(10x-2)=0

We get rid of parentheses

18x^2-16x+27x+(2x+3)(10x-2)-24=0

We multiply parentheses ..

18x^2+(+20x^2-4x+30x-6)-16x+27x-24=0

We add all the numbers together, and all the variables

18x^2+(+20x^2-4x+30x-6)+11x-24=0

We get rid of parentheses

18x^2+20x^2-4x+30x+11x-6-24=0

We add all the numbers together, and all the variables

38x^2+37x-30=0

a = 38; b = 37; c = -30;
Δ = b2-4ac
Δ = 372-4·38·(-30)
Δ = 5929
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5929}=77$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(37)-77}{2*38}=\frac{-114}{76}
=-1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(37)+77}{2*38}=\frac{40}{76}
=10/19
$