Evelyn G. answered • 11/06/19

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Two consecutive integers would be: 1st integer: x and 2nd integer: x+1, so with this knowledge we can then formulate the equation.

- 3x+(x+1)
^{2}=85 this is is a quadratic so we will need to treat it as such when solving. - 3x+(x+1)(x+1)=85
- 3x+x
^{2}+1x+1x+1=85 FOIL - x
^{2}+5x+1=85 combine like terms - x
^{2}+5x+1-85=85-85 subtract 85 from both sides to get into standard form Ax^{2}+Bx+C=0 - x
^{2}+5x-84=0 combine like terms - factors of -84:-1*84, -2*42, -3*28, -4*21, -6*14, -7*12
- which two factor sets subtract to get +5? -7*12 so 12-7 =5 check.
- (x-7)(x+12)=0 now we set each parenthesis = to 0 and solve to get two solutions.
- x-7=0 so x=7
- x+12=0 so x=-12
- now plug in each number back into the original equation to see which ones work. 3x+(x+1)
^{2}=85 - 3(7)+(7+1)
^{2}=85 gives 21+8^{2}= 21+64 =85 - 3(-12)+((-12)+1)
^{2}=85 gives -36+(-11)^{2}= -36+121 =85 - so since both solutions check out we can have x =7 or x =-12
- the first consecutive set would be 7 and 7+1 =8 solution 1: 7,8
- the second consecutive set would be -12 and -12+1=-11 solution 2: -12,-11