Quadratic equation form by factorization method

Ques2. TWO CIRCLES TOUCH EXTERNALLY.THE SUM OF THEIR AREA IS 130π SQUARE CM & DISTANCE BETWEEN THEIR CENTRES IS 14 cm .FIND THE RADII OF THE CIRCLES ?FIND ITS SOLUTION ONLY IN quadratic equation form by factorization method

let the radii be R and r:
R + r = 14 cm => r = 14 - R
A + a = 130π cm^2
πR^2 + πr^2 = 130π
R^2 + (14 - R)^2 = 130
R^2 + (196 - 28R + R^2) = 130
2R^2 - 28R + 196 - 130 = 0
R^2 - 14R + 33 = 0
(R - 3)(R - 11) = 0
R = 3 , 11 cm
r = 11 , 3 cm

Thank you and my best regards.

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Let the radius of first circle be X cm

Radius of the second circle will be (14 - X ) cm


Sum of the Area of the circles = pi X^2 + pi ( 14 - X)^2


= pi ( X^2 + 196 + X^2 - 28X) = 130 pi

So 2X^2 -28X + 66 = 0

X^2 - 14X + 33 = 0

( X - 11) ( X-3) = 0

X = 11 cm Or 3 cm ANSWER

A + B = 14..they touch, and the centres are 14 cm apart

pi*A^2 + pi*B^2 = 130*pi...sum of areas
A^2 + B^2 = 130
B = 14 - A
A^2 + (14 - A)^2 = 130
A^2 + 196 + A^2 - 28*A = 130
2*A^2 - 28*A + 66 = 0
A^2 - 14*A + 33 = 0
(A - 11)*(A - 3) = 0
A = 11 or A = 3
As B = 14 - A, the circles have radii 3 cm and 14 cm <<<

pi r^2 +pi R^2=130 pi
r^2 + R^2=130
and
r+R=14
so r^2+(14-r)^2=130
2r^2-28r+196=130
r^2-14r+33=0
(r-3)(r-11)=0
So radius of one circle is 3 and the other 11