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-8p^2-3p+4=0
a = -8; b = -3; c = +4;
Δ = b2-4ac
Δ = -32-4·(-8)·4
Δ = 137
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{137}}{2*-8}=\frac{3-\sqrt{137}}{-16} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{137}}{2*-8}=\frac{3+\sqrt{137}}{-16} $
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