k^2+7=56

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Solution for k^2+7=56 equation:



k^2+7=56
We move all terms to the left:
k^2+7-(56)=0
We add all the numbers together, and all the variables
k^2-49=0
a = 1; b = 0; c = -49;
Δ = b2-4ac
Δ = 02-4·1·(-49)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-14}{2*1}=\frac{-14}{2} =-7 $
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+14}{2*1}=\frac{14}{2} =7 $

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