(4x+1)-3(x-4)=(x+5)(x-5)-2

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Solution for (4x+1)-3(x-4)=(x+5)(x-5)-2 equation:



(4x+1)-3(x-4)=(x+5)(x-5)-2
We move all terms to the left:
(4x+1)-3(x-4)-((x+5)(x-5)-2)=0
We use the square of the difference formula
x^2+(4x+1)-3(x-4)+25=0
We multiply parentheses
x^2+(4x+1)-3x+12+25=0
We get rid of parentheses
x^2+4x-3x+1+12+25=0
We add all the numbers together, and all the variables
x^2+x+38=0
a = 1; b = 1; c = +38;
Δ = b2-4ac
Δ = 12-4·1·38
Δ = -151
Delta is less than zero, so there is no solution for the equation

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