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3z^2-5z^2+14=0
We add all the numbers together, and all the variables
-2z^2+14=0
a = -2; b = 0; c = +14;
Δ = b2-4ac
Δ = 02-4·(-2)·14
Δ = 112
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{7}}{2*-2}=\frac{0-4\sqrt{7}}{-4} =-\frac{4\sqrt{7}}{-4} =-\frac{\sqrt{7}}{-1} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{7}}{2*-2}=\frac{0+4\sqrt{7}}{-4} =\frac{4\sqrt{7}}{-4} =\frac{\sqrt{7}}{-1} $
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