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23x-8x^2+3=0
a = -8; b = 23; c = +3;
Δ = b2-4ac
Δ = 232-4·(-8)·3
Δ = 625
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{625}=25$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-25}{2*-8}=\frac{-48}{-16} =+3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+25}{2*-8}=\frac{2}{-16} =-1/8 $
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