-1+6k-((k-3)*(5k+3))=0

Solution for -1+6k-((k-3)*(5k+3))=0 equation:

-1+6k-((k-3)(5k+3))=0

We multiply parentheses ..

-((+5k^2+3k-15k-9))+6k-1=0


We calculate terms in parentheses: -((+5k^2+3k-15k-9)), so:

(+5k^2+3k-15k-9)

We get rid of parentheses

5k^2+3k-15k-9

We add all the numbers together, and all the variables

5k^2-12k-9

Back to the equation:

-(5k^2-12k-9)


We add all the numbers together, and all the variables

6k-(5k^2-12k-9)-1=0

We get rid of parentheses

-5k^2+6k+12k+9-1=0

We add all the numbers together, and all the variables

-5k^2+18k+8=0

a = -5; b = 18; c = +8;
Δ = b2-4ac
Δ = 182-4·(-5)·8
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-22}{2*-5}=\frac{-40}{-10}
=+4
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+22}{2*-5}=\frac{4}{-10}
=-2/5
$