(4k+5)(6k+10)=65

Solution for (4k+5)(6k+10)=65 equation:

(4k+5)(6k+10)=65

We move all terms to the left:

(4k+5)(6k+10)-(65)=0

We multiply parentheses ..

(+24k^2+40k+30k+50)-65=0

We get rid of parentheses

24k^2+40k+30k+50-65=0

We add all the numbers together, and all the variables

24k^2+70k-15=0

a = 24; b = 70; c = -15;
Δ = b2-4ac
Δ = 702-4·24·(-15)
Δ = 6340
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{6340}=\sqrt{4*1585}=\sqrt{4}*\sqrt{1585}=2\sqrt{1585}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(70)-2\sqrt{1585}}{2*24}=\frac{-70-2\sqrt{1585}}{48}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(70)+2\sqrt{1585}}{2*24}=\frac{-70+2\sqrt{1585}}{48}
$