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4(x+3)2x=x-8
We move all terms to the left:
4(x+3)2x-(x-8)=0
We multiply parentheses
8x^2+24x-(x-8)=0
We get rid of parentheses
8x^2+24x-x+8=0
We add all the numbers together, and all the variables
8x^2+23x+8=0
a = 8; b = 23; c = +8;
Δ = b2-4ac
Δ = 232-4·8·8
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{-(23)-\sqrt{273}}{2*8}=\frac{-23-\sqrt{273}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{273}}{2*8}=\frac{-23+\sqrt{273}}{16} $
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