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8x^2+23x+3=0
a = 8; b = 23; c = +3;
Δ = b2-4ac
Δ = 232-4·8·3
Δ = 433
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{433}}{2*8}=\frac{-23-\sqrt{433}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{433}}{2*8}=\frac{-23+\sqrt{433}}{16} $
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