-6c^2+7=0

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Solution for -6c^2+7=0 equation:



-6c^2+7=0
a = -6; b = 0; c = +7;
Δ = b2-4ac
Δ = 02-4·(-6)·7
Δ = 168
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{42}}{2*-6}=\frac{0-2\sqrt{42}}{-12} =-\frac{2\sqrt{42}}{-12} =-\frac{\sqrt{42}}{-6} $
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{42}}{2*-6}=\frac{0+2\sqrt{42}}{-12} =\frac{2\sqrt{42}}{-12} =\frac{\sqrt{42}}{-6} $

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