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540=(10+2x)(20+2x)
We move all terms to the left:
540-((10+2x)(20+2x))=0
We add all the numbers together, and all the variables
-((2x+10)(2x+20))+540=0
We multiply parentheses ..
-((+4x^2+40x+20x+200))+540=0
We calculate terms in parentheses: -((+4x^2+40x+20x+200)), so:We get rid of parentheses
(+4x^2+40x+20x+200)
We get rid of parentheses
4x^2+40x+20x+200
We add all the numbers together, and all the variables
4x^2+60x+200
Back to the equation:
-(4x^2+60x+200)
-4x^2-60x-200+540=0
We add all the numbers together, and all the variables
-4x^2-60x+340=0
a = -4; b = -60; c = +340;
Δ = b2-4ac
Δ = -602-4·(-4)·340
Δ = 9040
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{9040}=\sqrt{16*565}=\sqrt{16}*\sqrt{565}=4\sqrt{565}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-4\sqrt{565}}{2*-4}=\frac{60-4\sqrt{565}}{-8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+4\sqrt{565}}{2*-4}=\frac{60+4\sqrt{565}}{-8} $
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