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2x^2+23x-4.5=0
a = 2; b = 23; c = -4.5;
Δ = b2-4ac
Δ = 232-4·2·(-4.5)
Δ = 565
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{-(23)-\sqrt{565}}{2*2}=\frac{-23-\sqrt{565}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{565}}{2*2}=\frac{-23+\sqrt{565}}{4} $
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