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136=0.5x^2+0.5x
We move all terms to the left:
136-(0.5x^2+0.5x)=0
We get rid of parentheses
-0.5x^2-0.5x+136=0
a = -0.5; b = -0.5; c = +136;
Δ = b2-4ac
Δ = -0.52-4·(-0.5)·136
Δ = 272.25
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{-(-0.5)-\sqrt{272.25}}{2*-0.5}=\frac{0.5-\sqrt{272.25}}{-1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.5)+\sqrt{272.25}}{2*-0.5}=\frac{0.5+\sqrt{272.25}}{-1} $
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