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4x^2+23x+15=0
a = 4; b = 23; c = +15;
Δ = b2-4ac
Δ = 232-4·4·15
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-17}{2*4}=\frac{-40}{8} =-5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+17}{2*4}=\frac{-6}{8} =-3/4 $
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