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0=x(2x-3)
We move all terms to the left:
0-(x(2x-3))=0
We add all the numbers together, and all the variables
-(x(2x-3))=0
We calculate terms in parentheses: -(x(2x-3)), so:We get rid of parentheses
x(2x-3)
We multiply parentheses
2x^2-3x
Back to the equation:
-(2x^2-3x)
-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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-2}=\frac{-6}{-4} =1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-2}=\frac{0}{-4} =0 $
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