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4x^2+0x-8=0
We add all the numbers together, and all the variables
4x^2+x-8=0
a = 4; b = 1; c = -8;
Δ = b2-4ac
Δ = 12-4·4·(-8)
Δ = 129
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{-(1)-\sqrt{129}}{2*4}=\frac{-1-\sqrt{129}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{129}}{2*4}=\frac{-1+\sqrt{129}}{8} $
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