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2x^2-5x-16=0
a = 2; b = -5; c = -16;
Δ = b2-4ac
Δ = -52-4·2·(-16)
Δ = 153
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{153}=\sqrt{9*17}=\sqrt{9}*\sqrt{17}=3\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-3\sqrt{17}}{2*2}=\frac{5-3\sqrt{17}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+3\sqrt{17}}{2*2}=\frac{5+3\sqrt{17}}{4} $
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