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2x^2-26x+15=0
a = 2; b = -26; c = +15;
Δ = b2-4ac
Δ = -262-4·2·15
Δ = 556
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{556}=\sqrt{4*139}=\sqrt{4}*\sqrt{139}=2\sqrt{139}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2\sqrt{139}}{2*2}=\frac{26-2\sqrt{139}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2\sqrt{139}}{2*2}=\frac{26+2\sqrt{139}}{4} $
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