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3z^2=26
We move all terms to the left:
3z^2-(26)=0
a = 3; b = 0; c = -26;
Δ = b2-4ac
Δ = 02-4·3·(-26)
Δ = 312
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{312}=\sqrt{4*78}=\sqrt{4}*\sqrt{78}=2\sqrt{78}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{78}}{2*3}=\frac{0-2\sqrt{78}}{6} =-\frac{2\sqrt{78}}{6} =-\frac{\sqrt{78}}{3} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{78}}{2*3}=\frac{0+2\sqrt{78}}{6} =\frac{2\sqrt{78}}{6} =\frac{\sqrt{78}}{3} $
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