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=(30+2A)(20+2A)
We move all terms to the left:
-((30+2A)(20+2A))=0
We add all the numbers together, and all the variables
-((2A+30)(2A+20))=0
We multiply parentheses ..
-((+4A^2+40A+60A+600))=0
We calculate terms in parentheses: -((+4A^2+40A+60A+600)), so:We get rid of parentheses
(+4A^2+40A+60A+600)
We get rid of parentheses
4A^2+40A+60A+600
We add all the numbers together, and all the variables
4A^2+100A+600
Back to the equation:
-(4A^2+100A+600)
-4A^2-100A-600=0
a = -4; b = -100; c = -600;
Δ = b2-4ac
Δ = -1002-4·(-4)·(-600)
Δ = 400
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{400}=20$$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-100)-20}{2*-4}=\frac{80}{-8} =-10 $$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+20}{2*-4}=\frac{120}{-8} =-15 $
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