0=(3x+4)(4x-5)

Solution for 0=(3x+4)(4x-5) equation:

0=(3x+4)(4x-5)

We move all terms to the left:

0-((3x+4)(4x-5))=0

We add all the numbers together, and all the variables

-((3x+4)(4x-5))=0

We multiply parentheses ..

-((+12x^2-15x+16x-20))=0


We calculate terms in parentheses: -((+12x^2-15x+16x-20)), so:

(+12x^2-15x+16x-20)

We get rid of parentheses

12x^2-15x+16x-20

We add all the numbers together, and all the variables

12x^2+x-20

Back to the equation:

-(12x^2+x-20)


We get rid of parentheses

-12x^2-x+20=0

We add all the numbers together, and all the variables

-12x^2-1x+20=0

a = -12; b = -1; c = +20;
Δ = b2-4ac
Δ = -12-4·(-12)·20
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-31}{2*-12}=\frac{-30}{-24}
=1+1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+31}{2*-12}=\frac{32}{-24}
=-1+1/3
$