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24x^2+26x+5=0
a = 24; b = 26; c = +5;
Δ = b2-4ac
Δ = 262-4·24·5
Δ = 196
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{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-14}{2*24}=\frac{-40}{48} =-5/6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+14}{2*24}=\frac{-12}{48} =-1/4 $
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