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4x^2+23x+8=0
a = 4; b = 23; c = +8;
Δ = b2-4ac
Δ = 232-4·4·8
Δ = 401
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{401}}{2*4}=\frac{-23-\sqrt{401}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{401}}{2*4}=\frac{-23+\sqrt{401}}{8} $
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