X^2+13x+40=(x+5)

Solution for X^2+13x+40=(x+5) equation:

X^2+13X+40=(X+5)

We move all terms to the left:

X^2+13X+40-((X+5))=0


We calculate terms in parentheses: -((X+5)), so:

(X+5)

We get rid of parentheses

X+5

Back to the equation:

-(X+5)


We get rid of parentheses

X^2+13X-X-5+40=0

We add all the numbers together, and all the variables

X^2+12X+35=0

a = 1; b = 12; c = +35;
Δ = b2-4ac
Δ = 122-4·1·35
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2}{2*1}=\frac{-14}{2}
=-7
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2}{2*1}=\frac{-10}{2}
=-5
$