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x=-6x^2=5x-3
We move all terms to the left:
x-(-6x^2)=0
We get rid of parentheses
6x^2+x=0
a = 6; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·6·0
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*6}=\frac{-2}{12} =-1/6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*6}=\frac{0}{12} =0 $
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