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