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24x^2+22x+3=0
a = 24; b = 22; c = +3;
Δ = b2-4ac
Δ = 222-4·24·3
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-14}{2*24}=\frac{-36}{48} =-3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+14}{2*24}=\frac{-8}{48} =-1/6 $
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