13+2k=3k+4k(k-3)

Solution for 13+2k=3k+4k(k-3) equation:

13+2k=3k+4k(k-3)

We move all terms to the left:

13+2k-(3k+4k(k-3))=0


We calculate terms in parentheses: -(3k+4k(k-3)), so:

3k+4k(k-3)

We multiply parentheses

4k^2+3k-12k

We add all the numbers together, and all the variables

4k^2-9k

Back to the equation:

-(4k^2-9k)


We get rid of parentheses

-4k^2+2k+9k+13=0

We add all the numbers together, and all the variables

-4k^2+11k+13=0

a = -4; b = 11; c = +13;
Δ = b2-4ac
Δ = 112-4·(-4)·13
Δ = 329
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{329}}{2*-4}=\frac{-11-\sqrt{329}}{-8}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{329}}{2*-4}=\frac{-11+\sqrt{329}}{-8}
$