(3k-8)(k+3)=-17

Solution for (3k-8)(k+3)=-17 equation:

(3k-8)(k+3)=-17

We move all terms to the left:

(3k-8)(k+3)-(-17)=0

We add all the numbers together, and all the variables

(3k-8)(k+3)+17=0

We multiply parentheses ..

(+3k^2+9k-8k-24)+17=0

We get rid of parentheses

3k^2+9k-8k-24+17=0

We add all the numbers together, and all the variables

3k^2+k-7=0

a = 3; b = 1; c = -7;
Δ = b2-4ac
Δ = 12-4·3·(-7)
Δ = 85
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{-(1)-\sqrt{85}}{2*3}=\frac{-1-\sqrt{85}}{6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{85}}{2*3}=\frac{-1+\sqrt{85}}{6}
$