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16x^2-20x-62=0
a = 16; b = -20; c = -62;
Δ = b2-4ac
Δ = -202-4·16·(-62)
Δ = 4368
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4368}=\sqrt{16*273}=\sqrt{16}*\sqrt{273}=4\sqrt{273}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{273}}{2*16}=\frac{20-4\sqrt{273}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{273}}{2*16}=\frac{20+4\sqrt{273}}{32} $
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