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-2x^2=3x
We move all terms to the left:
-2x^2-(3x)=0
a = -2; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·(-2)·0
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*-2}=\frac{0}{-4} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*-2}=\frac{6}{-4} =-1+1/2 $
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