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20x^2+43x+15=0
a = 20; b = 43; c = +15;
Δ = b2-4ac
Δ = 432-4·20·15
Δ = 649
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-\sqrt{649}}{2*20}=\frac{-43-\sqrt{649}}{40} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+\sqrt{649}}{2*20}=\frac{-43+\sqrt{649}}{40} $
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