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28x^2+25x+3=0
a = 28; b = 25; c = +3;
Δ = b2-4ac
Δ = 252-4·28·3
Δ = 289
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{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-17}{2*28}=\frac{-42}{56} =-3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+17}{2*28}=\frac{-8}{56} =-1/7 $
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