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17=0.5x^2+3x
We move all terms to the left:
17-(0.5x^2+3x)=0
We get rid of parentheses
-0.5x^2-3x+17=0
a = -0.5; b = -3; c = +17;
Δ = b2-4ac
Δ = -32-4·(-0.5)·17
Δ = 43
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{-(-3)-\sqrt{43}}{2*-0.5}=\frac{3-\sqrt{43}}{-1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{43}}{2*-0.5}=\frac{3+\sqrt{43}}{-1} $
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