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z^2+14z+46=0
a = 1; b = 14; c = +46;
Δ = b2-4ac
Δ = 142-4·1·46
Δ = 12
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{3}}{2*1}=\frac{-14-2\sqrt{3}}{2} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{3}}{2*1}=\frac{-14+2\sqrt{3}}{2} $
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