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7z(2)=56
We move all terms to the left:
7z(2)-(56)=0
We add all the numbers together, and all the variables
7z^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:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1568}=\sqrt{784*2}=\sqrt{784}*\sqrt{2}=28\sqrt{2}$$z_{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} $$z_{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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