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Divide 20 into two parts such that three times the square of one part exceeds the other part by 10 in quadratic equations

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Divide 20 into two parts such that three times the square of one part exceeds the other part by 10 in quadratic equations

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Let two parts be x and y.thereforex+y=20 ⇒ y=20-x

Then we have 3x2-(20-x)=10

3x2+x-30=0

3x2+10x-9x-30=0

x(3x+10)-3(3x+10)=0

(x-3)(3x+10)=0

x=3 is the solution.

The two parts are 3 and 17 respectively.

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