(k+4)(k+7)=k+28

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Solution for (k+4)(k+7)=k+28 equation:



(k+4)(k+7)=k+28
We move all terms to the left:
(k+4)(k+7)-(k+28)=0
We get rid of parentheses
(k+4)(k+7)-k-28=0
We multiply parentheses ..
(+k^2+7k+4k+28)-k-28=0
We add all the numbers together, and all the variables
(+k^2+7k+4k+28)-1k-28=0
We get rid of parentheses
k^2+7k+4k-1k+28-28=0
We add all the numbers together, and all the variables
k^2+10k=0
a = 1; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·1·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*1}=\frac{-20}{2} =-10 $
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*1}=\frac{0}{2} =0 $

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