7w^2+28=84

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Solution for 7w^2+28=84 equation:



7w^2+28=84
We move all terms to the left:
7w^2+28-(84)=0
We add all the numbers together, and all the variables
7w^2-56=0
a = 7; b = 0; c = -56;
Δ = b2-4ac
Δ = 02-4·7·(-56)
Δ = 1568
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{1568}=\sqrt{784*2}=\sqrt{784}*\sqrt{2}=28\sqrt{2}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-28\sqrt{2}}{2*7}=\frac{0-28\sqrt{2}}{14} =-\frac{28\sqrt{2}}{14} =-2\sqrt{2} $
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+28\sqrt{2}}{2*7}=\frac{0+28\sqrt{2}}{14} =\frac{28\sqrt{2}}{14} =2\sqrt{2} $

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