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x^2+16x-62=0
a = 1; b = 16; c = -62;
Δ = b2-4ac
Δ = 162-4·1·(-62)
Δ = 504
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-6\sqrt{14}}{2*1}=\frac{-16-6\sqrt{14}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+6\sqrt{14}}{2*1}=\frac{-16+6\sqrt{14}}{2} $
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