(u+4)-(u-3)(u+3)=3(u-10)

Solution for (u+4)-(u-3)(u+3)=3(u-10) equation:

(u+4)-(u-3)(u+3)=3(u-10)

We move all terms to the left:

(u+4)-(u-3)(u+3)-(3(u-10))=0

We use the square of the difference formula

u^2+(u+4)-(3(u-10))+9=0

We get rid of parentheses

u^2+u-(3(u-10))+4+9=0


We calculate terms in parentheses: -(3(u-10)), so:

3(u-10)

We multiply parentheses

3u-30

Back to the equation:

-(3u-30)


We add all the numbers together, and all the variables

u^2+u-(3u-30)+13=0

We get rid of parentheses

u^2+u-3u+30+13=0

We add all the numbers together, and all the variables

u^2-2u+43=0

a = 1; b = -2; c = +43;
Δ = b2-4ac
Δ = -22-4·1·43
Δ = -168
Delta is less than zero, so there is no solution for the equation