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7z^2-14z-7=0
a = 7; b = -14; c = -7;
Δ = b2-4ac
Δ = -142-4·7·(-7)
Δ = 392
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{392}=\sqrt{196*2}=\sqrt{196}*\sqrt{2}=14\sqrt{2}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-14\sqrt{2}}{2*7}=\frac{14-14\sqrt{2}}{14} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+14\sqrt{2}}{2*7}=\frac{14+14\sqrt{2}}{14} $
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