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5x^2-16x-14=0
a = 5; b = -16; c = -14;
Δ = b2-4ac
Δ = -162-4·5·(-14)
Δ = 536
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{536}=\sqrt{4*134}=\sqrt{4}*\sqrt{134}=2\sqrt{134}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{134}}{2*5}=\frac{16-2\sqrt{134}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{134}}{2*5}=\frac{16+2\sqrt{134}}{10} $
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