(6x-3)2-4(x-3)(x+3)=27x2+45

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Solution for (6x-3)2-4(x-3)(x+3)=27x2+45 equation: 



(6x-3)2-4(x-3)(x+3)=27x^2+45

We move all terms to the left:

(6x-3)2-4(x-3)(x+3)-(27x^2+45)=0

We use the square of the difference formula

x^2+(6x-3)2-(27x^2+45)+9=0

We multiply parentheses

x^2+12x-(27x^2+45)-6+9=0

We get rid of parentheses

x^2-27x^2+12x-45-6+9=0

We add all the numbers together, and all the variables

-26x^2+12x-42=0

a = -26; b = 12; c = -42;
Δ = b2-4ac
Δ = 122-4·(-26)·(-42)
Δ = -4224
Delta is less than zero, so there is no solution for the equation

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