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(2x^2-8x)-5=0
We get rid of parentheses
2x^2-8x-5=0
a = 2; b = -8; c = -5;
Δ = b2-4ac
Δ = -82-4·2·(-5)
Δ = 104
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{104}=\sqrt{4*26}=\sqrt{4}*\sqrt{26}=2\sqrt{26}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{26}}{2*2}=\frac{8-2\sqrt{26}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{26}}{2*2}=\frac{8+2\sqrt{26}}{4} $
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