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5x^2+20x=2x+8
We move all terms to the left:
5x^2+20x-(2x+8)=0
We get rid of parentheses
5x^2+20x-2x-8=0
We add all the numbers together, and all the variables
5x^2+18x-8=0
a = 5; b = 18; c = -8;
Δ = b2-4ac
Δ = 182-4·5·(-8)
Δ = 484
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{484}=22$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-22}{2*5}=\frac{-40}{10} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+22}{2*5}=\frac{4}{10} =2/5 $
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