10+(2x+1)^2/3=13

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Solution for 10+(2x+1)^2/3=13 equation: 


x in (-oo:+oo)

((2*x+1)^2)/3+10 = 13 // - 13

((2*x+1)^2)/3-13+10 = 0

((2*x+1)^2)/3+(-13*3)/3+(3*10)/3 = 0

(2*x+1)^2-13*3+3*10 = 0

4*x^2+4*x-38+30 = 0

4*x^2+4*x-8 = 0

4*x^2+4*x-8 = 0

4*(x^2+x-2) = 0

x^2+x-2 = 0

DELTA = 1^2-(-2*1*4)

DELTA = 9

DELTA > 0

x = (9^(1/2)-1)/(1*2) or x = (-9^(1/2)-1)/(1*2)

x = 1 or x = -2

4*(x+2)*(x-1) = 0

(4*(x+2)*(x-1))/3 = 0

(4*(x+2)*(x-1))/3 = 0 // * 3

4*(x+2)*(x-1) = 0

( x+2 )

x+2 = 0 // - 2

x = -2

( x-1 )

x-1 = 0 // + 1

x = 1

x in { -2, 1 }

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