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4x^2+43x+115=0
a = 4; b = 43; c = +115;
Δ = b2-4ac
Δ = 432-4·4·115
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-3}{2*4}=\frac{-46}{8} =-5+3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+3}{2*4}=\frac{-40}{8} =-5 $
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