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=A(3A-5)
We move all terms to the left:
-(A(3A-5))=0
We calculate terms in parentheses: -(A(3A-5)), so:We get rid of parentheses
A(3A-5)
We multiply parentheses
3A^2-5A
Back to the equation:
-(3A^2-5A)
-3A^2+5A=0
a = -3; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·(-3)·0
Δ = 25
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{25}=5$$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*-3}=\frac{-10}{-6} =1+2/3 $$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*-3}=\frac{0}{-6} =0 $
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