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x^2-16x-53=0
a = 1; b = -16; c = -53;
Δ = b2-4ac
Δ = -162-4·1·(-53)
Δ = 468
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-6\sqrt{13}}{2*1}=\frac{16-6\sqrt{13}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+6\sqrt{13}}{2*1}=\frac{16+6\sqrt{13}}{2} $
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