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5x^2+23x+12=0
a = 5; b = 23; c = +12;
Δ = b2-4ac
Δ = 232-4·5·12
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-17}{2*5}=\frac{-40}{10} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+17}{2*5}=\frac{-6}{10} =-3/5 $
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