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(4k+5)(6k+10)=115
We move all terms to the left:
(4k+5)(6k+10)-(115)=0
We multiply parentheses ..
(+24k^2+40k+30k+50)-115=0
We get rid of parentheses
24k^2+40k+30k+50-115=0
We add all the numbers together, and all the variables
24k^2+70k-65=0
a = 24; b = 70; c = -65;
Δ = b2-4ac
Δ = 702-4·24·(-65)
Δ = 11140
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{11140}=\sqrt{4*2785}=\sqrt{4}*\sqrt{2785}=2\sqrt{2785}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(70)-2\sqrt{2785}}{2*24}=\frac{-70-2\sqrt{2785}}{48} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(70)+2\sqrt{2785}}{2*24}=\frac{-70+2\sqrt{2785}}{48} $
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