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8x^2-9x+3-5-4=0
We add all the numbers together, and all the variables
8x^2-9x-6=0
a = 8; b = -9; c = -6;
Δ = b2-4ac
Δ = -92-4·8·(-6)
Δ = 273
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{-(-9)-\sqrt{273}}{2*8}=\frac{9-\sqrt{273}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{273}}{2*8}=\frac{9+\sqrt{273}}{16} $
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