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=6A^2+3A
We move all terms to the left:
-(6A^2+3A)=0
We get rid of parentheses
-6A^2-3A=0
a = -6; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·(-6)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{9}=3$$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*-6}=\frac{0}{-12} =0 $$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*-6}=\frac{6}{-12} =-1/2 $
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