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0.1x^2+2x+10=20
We move all terms to the left:
0.1x^2+2x+10-(20)=0
We add all the numbers together, and all the variables
0.1x^2+2x-10=0
a = 0.1; b = 2; c = -10;
Δ = b2-4ac
Δ = 22-4·0.1·(-10)
Δ = 8
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{2}}{2*0.1}=\frac{-2-2\sqrt{2}}{0.2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{2}}{2*0.1}=\frac{-2+2\sqrt{2}}{0.2} $
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