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how do i solve -3r+3r=-6(7+7r)-7(4-7r)
To answer a quadratic like this, st everything equal to zero. 5r^2+12r=-4r^2-4 9r^2+12r+4=0 Ask yourself what two numbers add to get 12 that multiply to get 9*4=36 1*36 no 2*18 no 3*12 no 4*9 no 6*6 yes (3r+2)(3r+2) The first and last multiply to get 6 and the outside and inside multiply to get 6. That's (3r+)^2
6p + 12q + 18r = 6 (p + 2q + 3r)
3.5
6, remainder 21