(x=4)(x^2-x+3)=0

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



(x=4)(x^2-x+3)=0

We move all terms to the left:

(x-(4)(x^2-x+3))=0



We calculate terms in parentheses: +(x-4(x^2-x+3)), so:

x-4(x^2-x+3)

We multiply parentheses

-4x^2+x+4x-12

We add all the numbers together, and all the variables

-4x^2+5x-12

Back to the equation:

+(-4x^2+5x-12)



We get rid of parentheses

-4x^2+5x-12=0

a = -4; b = 5; c = -12;
Δ = b2-4ac
Δ = 52-4·(-4)·(-12)
Δ = -167
Delta is less than zero, so there is no solution for the equation

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