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26x^2+23x+5=0
a = 26; b = 23; c = +5;
Δ = b2-4ac
Δ = 232-4·26·5
Δ = 9
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-3}{2*26}=\frac{-26}{52} =-1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+3}{2*26}=\frac{-20}{52} =-5/13 $
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