(u2+(-24)=u+35

Solution for (u2+(-24)=u+35 equation:

(u2+(-24)=u+35

We move all terms to the left:

(u2+(-24)-(u+35)=0


We calculate terms in parentheses: +(u2+(-24)-(u+35), so:

u2+(-24)-(u+35

determiningTheFunctionDomain
u2-(u+35+(-24)

We add all the numbers together, and all the variables

u^2-(u+35+(-24)


We calculate terms in parentheses: -(u+35+(-24), so:

u+35+(-24

We add all the numbers together, and all the variables

u

Back to the equation:

-(u)


We add all the numbers together, and all the variables

u^2-1u

Back to the equation:

+(u^2-1u)


We get rid of parentheses

u^2-1u=0

a = 1; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·1·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*1}=\frac{0}{2}
=0
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*1}=\frac{2}{2}
=1
$