(2x-3)(6x-4)+20=0

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Solution for (2x-3)(6x-4)+20=0 equation: 



(2x-3)(6x-4)+20=0

We multiply parentheses ..

(+12x^2-8x-18x+12)+20=0

We get rid of parentheses

12x^2-8x-18x+12+20=0

We add all the numbers together, and all the variables

12x^2-26x+32=0

a = 12; b = -26; c = +32;
Δ = b2-4ac
Δ = -262-4·12·32
Δ = -860
Delta is less than zero, so there is no solution for the equation

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