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0.5x^2=-6x+20
We move all terms to the left:
0.5x^2-(-6x+20)=0
We get rid of parentheses
0.5x^2+6x-20=0
a = 0.5; b = 6; c = -20;
Δ = b2-4ac
Δ = 62-4·0.5·(-20)
Δ = 76
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{19}}{2*0.5}=\frac{-6-2\sqrt{19}}{1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{19}}{2*0.5}=\frac{-6+2\sqrt{19}}{1} $
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