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=2A^2-3A+3A^2+2A-3
We move all terms to the left:
-(2A^2-3A+3A^2+2A-3)=0
We get rid of parentheses
-2A^2-3A^2+3A-2A+3=0
We add all the numbers together, and all the variables
-5A^2+A+3=0
a = -5; b = 1; c = +3;
Δ = b2-4ac
Δ = 12-4·(-5)·3
Δ = 61
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}$$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{61}}{2*-5}=\frac{-1-\sqrt{61}}{-10} $$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{61}}{2*-5}=\frac{-1+\sqrt{61}}{-10} $
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