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0.5x^2-20x=88
We move all terms to the left:
0.5x^2-20x-(88)=0
a = 0.5; b = -20; c = -88;
Δ = b2-4ac
Δ = -202-4·0.5·(-88)
Δ = 576
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{576}=24$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-24}{2*0.5}=\frac{-4}{1} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+24}{2*0.5}=\frac{44}{1} =44 $
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