(2x+5)(6x-4)=0

Solution for (2x+5)(6x-4)=0 equation:

(2x+5)(6x-4)=0

We multiply parentheses ..

(+12x^2-8x+30x-20)=0

We get rid of parentheses

12x^2-8x+30x-20=0

We add all the numbers together, and all the variables

12x^2+22x-20=0

a = 12; b = 22; c = -20;
Δ = b2-4ac
Δ = 222-4·12·(-20)
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{-(22)-38}{2*12}=\frac{-60}{24}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+38}{2*12}=\frac{16}{24}
=2/3
$